\displaystyle \text{MULTIPLE CHOICE QUESTIONS (MCQs)}

\displaystyle \textbf{Question 1: }\text{If the equation of a circle is }\lambda x^2+(2\lambda-3)y^2
\displaystyle -4x+6y-1=0,\text{ then the coordinates of centre are}
\displaystyle \text{(a) }(4/3,-1)\qquad\text{(b) }(2/3,-1)
\displaystyle \text{(c) }(-2/3,1)\qquad\text{(d) }(2/3,1)
\displaystyle \text{Answer:}
\displaystyle \text{For the equation to represent a circle, coefficients of }x^2\text{ and }y^2
\displaystyle \text{must be equal.}
\displaystyle \therefore \lambda=2\lambda-3\Rightarrow\lambda=3.
\displaystyle \therefore 3x^2+3y^2-4x+6y-1=0.
\displaystyle \Rightarrow x^2+y^2-\frac43x+2y-\frac13=0.
\displaystyle \text{Comparing with }x^2+y^2+2gx+2fy+c=0,
\displaystyle 2g=-\frac43\Rightarrow g=-\frac23,\qquad 2f=2\Rightarrow f=1.
\displaystyle \therefore \text{Centre}=(-g,-f)=\left(\frac23,-1\right).
\displaystyle \therefore \text{Correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }4x^2+2\lambda xy+4y^2+2(\lambda-4)x+12y-59=0
\displaystyle \text{is the equation of a circle, then its radius is}
\displaystyle \text{(a) }3\sqrt{2}\qquad\text{(b) }2\sqrt{3}\qquad \text{(c) }2\sqrt{2}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{For the given equation to represent a circle, the coefficient of }xy\text{ must be zero.}
\displaystyle \therefore 2\lambda=0\Rightarrow\lambda=0.
\displaystyle \text{Thus, }4x^2+4y^2-8x+12y-59=0.
\displaystyle \text{Dividing by }4,
\displaystyle x^2+y^2-2x+3y-\frac{59}{4}=0.
\displaystyle \text{Comparing with }x^2+y^2+2gx+2fy+c=0,
\displaystyle g=-1,\qquad f=\frac{3}{2},\qquad c=-\frac{59}{4}.
\displaystyle r=\sqrt{g^2+f^2-c}.
\displaystyle =\sqrt{1+\frac{9}{4}+\frac{59}{4}}
\displaystyle =\sqrt{\frac{72}{4}}=\sqrt{18}=3\sqrt{2}.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The equation }x^2+y^2+2x-4y+5=0\text{ represents}
\displaystyle \text{(a) a point}\qquad\text{(b) a pair of straight lines}
\displaystyle \text{(c) a circle of non-zero radius}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle x^2+y^2+2x-4y+5=0.
\displaystyle (x+1)^2-1+(y-2)^2-4+5=0.
\displaystyle \Rightarrow (x+1)^2+(y-2)^2=0.
\displaystyle \therefore\text{ The circle has centre }(-1,2)\text{ and radius }0.
\displaystyle \therefore\text{ It represents a point.}
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If the equation }(4a-3)x^2+ay^2+6x-2y+2=0
\displaystyle \text{represents a circle, then its centre is}
\displaystyle \text{(a) }(3,-1)\qquad\text{(b) }(3,1)\qquad\text{(c) }(-3,1)
\displaystyle \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{For a circle, coefficients of }x^2\text{ and }y^2\text{ must be equal.}
\displaystyle 4a-3=a\Rightarrow 3a=3\Rightarrow a=1.
\displaystyle \therefore x^2+y^2+6x-2y+2=0.
\displaystyle \text{Comparing with }x^2+y^2+2gx+2fy+c=0,
\displaystyle 2g=6\Rightarrow g=3,\qquad 2f=-2\Rightarrow f=-1.
\displaystyle \therefore\text{ Centre}=(-g,-f)=(-3,1).
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The radius of the circle represented by the equation}
\displaystyle 3x^2+3y^2+\lambda xy+9x+(\lambda-6)y+3=0\text{ is}
\displaystyle \text{(a) }\frac{3}{2}\qquad\text{(b) }\frac{\sqrt{17}}{2}\qquad\text{(c) }\frac{2}{3}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{For the equation to represent a circle, the coefficient of }xy\text{ must be zero.}
\displaystyle \therefore\lambda=0.
\displaystyle 3x^2+3y^2+9x-6y+3=0.
\displaystyle \Rightarrow x^2+y^2+3x-2y+1=0.
\displaystyle \text{Comparing with }x^2+y^2+2gx+2fy+c=0,
\displaystyle g=\frac{3}{2},\qquad f=-1,\qquad c=1.
\displaystyle r=\sqrt{g^2+f^2-c}
\displaystyle =\sqrt{\frac{9}{4}+1-1}=\frac{3}{2}.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The number of integral values of }\lambda\text{ for which the equation}
\displaystyle x^2+y^2+\lambda x+(1-\lambda)y+5=0\text{ is the equation of a}
\displaystyle \text{circle whose radius cannot exceed }5,\text{ is}
\displaystyle \text{(a) }14\qquad\text{(b) }18\qquad\text{(c) }16\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Comparing with }x^2+y^2+2gx+2fy+c=0,
\displaystyle g=\frac{\lambda}{2},\qquad f=\frac{1-\lambda}{2},\qquad c=5.
\displaystyle r^2=g^2+f^2-c
\displaystyle =\frac{\lambda^2}{4}+\frac{(1-\lambda)^2}{4}-5.
\displaystyle \text{Since the radius cannot exceed }5,\quad r^2\leq25.
\displaystyle \frac{\lambda^2+(1-\lambda)^2}{4}-5\leq25.
\displaystyle \Rightarrow\lambda^2+(1-\lambda)^2\leq120.
\displaystyle \Rightarrow2\lambda^2-2\lambda-119\leq0.
\displaystyle \Rightarrow\frac{1-\sqrt{239}}{2}\leq\lambda\leq\frac{1+\sqrt{239}}{2}.
\displaystyle \text{Since }15<\sqrt{239}<16,
\displaystyle -7\leq\lambda\leq8.
\displaystyle \therefore\lambda=-7,-6,-5,\ldots,7,8.
\displaystyle \therefore\text{ Number of integral values}=16.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The equation of the circle passing through the point }(1,1)
\displaystyle \text{and having two diameters along the pair of lines}
\displaystyle x^2-y^2-2x+4y-3=0,\text{ is}
\displaystyle \text{(a) }x^2+y^2-2x-4y+4=0
\displaystyle \text{(b) }x^2+y^2+2x+4y-4=0
\displaystyle \text{(c) }x^2+y^2-2x+4y+4=0\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle x^2-y^2-2x+4y-3=0
\displaystyle \Rightarrow (x-1)^2-(y-2)^2=0.
\displaystyle \Rightarrow (x-y+1)(x+y-3)=0.
\displaystyle \text{Thus, the two diameters are }x-y+1=0\text{ and }x+y-3=0.
\displaystyle \text{Their point of intersection gives the centre of the circle.}
\displaystyle x-y+1=0,\qquad x+y-3=0
\displaystyle \Rightarrow x=1,\qquad y=2.
\displaystyle \therefore\text{ Centre}=(1,2).
\displaystyle \text{The circle passes through }(1,1).
\displaystyle \therefore r^2=(1-1)^2+(1-2)^2=1.
\displaystyle \therefore (x-1)^2+(y-2)^2=1.
\displaystyle \Rightarrow x^2+y^2-2x-4y+4=0.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If the centroid of an equilateral triangle is }(1,1)\text{ and its}
\displaystyle \text{one vertex is }(-1,2),\text{ then the equation of its circumcircle is}
\displaystyle \text{(a) }x^2+y^2-2x-2y-3=0\qquad\text{(b) }x^2+y^2+2x-2y-3=0
\displaystyle \text{(c) }x^2+y^2+2x+2y-3=0\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{In an equilateral triangle, the centroid and circumcentre coincide.}
\displaystyle \therefore\text{ Centre of the circumcircle}=(1,1).
\displaystyle \text{Since }(-1,2)\text{ is a vertex,}
\displaystyle r^2=(-1-1)^2+(2-1)^2=4+1=5.
\displaystyle \therefore (x-1)^2+(y-1)^2=5.
\displaystyle \Rightarrow x^2+y^2-2x-2y-3=0.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If the point }(2,k)\text{ lies outside the circles}
\displaystyle x^2+y^2+x-2y-14=0\text{ and }x^2+y^2=13,\text{ then }k\text{ lies in the interval}
\displaystyle \text{(a) }(-3,-2)\cup(3,4)\qquad\text{(b) }-3,4
\displaystyle \text{(c) }(-\infty,-3)\cup(4,\infty)\qquad\text{(d) }(-\infty,-2)\cup(3,\infty)
\displaystyle \text{Answer:}
\displaystyle \text{For }x^2+y^2+x-2y-14=0,\text{ the point }(2,k)\text{ lies outside if}
\displaystyle 2^2+k^2+2-2k-14>0.
\displaystyle \Rightarrow k^2-2k-8>0
\displaystyle \Rightarrow (k-4)(k+2)>0.
\displaystyle \Rightarrow k<-2\text{ or }k>4.
\displaystyle \text{For }x^2+y^2=13,\text{ the point }(2,k)\text{ lies outside if}
\displaystyle 2^2+k^2>13.
\displaystyle \Rightarrow k^2>9\Rightarrow k<-3\text{ or }k>3.
\displaystyle \text{For the point to lie outside both circles, we take the intersection.}
\displaystyle [(-\infty,-2)\cup(4,\infty)]\cap[(-\infty,-3)\cup(3,\infty)]
\displaystyle =(-\infty,-3)\cup(4,\infty).
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If the point }(\lambda,\lambda+1)\text{ lies inside the region}
\displaystyle \text{bounded by the curve }x=\sqrt{25-y^2}\text{ and }y\text{-axis, then }\lambda\text{ belongs to the}
\displaystyle \text{interval}
\displaystyle \text{(a) }(-1,3)\qquad\text{(b) }(-4,3)
\displaystyle \text{(c) }(-\infty,-4)\cup(3,\infty)\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle x=\sqrt{25-y^2}\Rightarrow x^2+y^2=25,\qquad x\geq0.
\displaystyle \text{For }(\lambda,\lambda+1)\text{ to lie inside the region,}
\displaystyle \lambda\geq0\quad\text{and}\quad\lambda^2+(\lambda+1)^2<25.
\displaystyle \Rightarrow2\lambda^2+2\lambda-24<0.
\displaystyle \Rightarrow\lambda^2+\lambda-12<0.
\displaystyle \Rightarrow(\lambda+4)(\lambda-3)<0.
\displaystyle \Rightarrow-4<\lambda<3.
\displaystyle \text{Also, }\lambda\geq0\Rightarrow0\leq\lambda<3.
\displaystyle \text{Since }[0,3)\subset(-1,3),\text{ the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{The equation of the incircle formed by the coordinate axes}
\displaystyle \text{and the line }4x+3y=6\text{ is}
\displaystyle \text{(a) }x^2+y^2-6x-6y+9=0
\displaystyle \text{(b) }4(x^2+y^2-x-y)+1=0
\displaystyle \text{(c) }4(x^2+y^2+x+y)+1=0\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Since the circle touches both coordinate axes, let its centre be }(r,r).
\displaystyle \text{Its distance from }4x+3y-6=0\text{ is also }r.
\displaystyle \frac{|4r+3r-6|}{\sqrt{4^2+3^2}}=r.
\displaystyle \frac{6-7r}{5}=r\Rightarrow6-7r=5r\Rightarrow r=\frac12.
\displaystyle \therefore\text{ Centre}=\left(\frac12,\frac12\right),\qquad r=\frac12.
\displaystyle \therefore\left(x-\frac12\right)^2+\left(y-\frac12\right)^2=\frac14.
\displaystyle \Rightarrow x^2+y^2-x-y+\frac14=0.
\displaystyle \Rightarrow4(x^2+y^2-x-y)+1=0.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If the circles }x^2+y^2=9\text{ and}
\displaystyle x^2+y^2+8y+c=0\text{ touch each other, then }c\text{ is equal to}
\displaystyle \text{(a) }15\qquad\text{(b) }-15\qquad\text{(c) }16\qquad\text{(d) }-16
\displaystyle \text{Answer:}
\displaystyle x^2+y^2=9\text{ has centre }C_1=(0,0)\text{ and radius }r_1=3.
\displaystyle x^2+y^2+8y+c=0
\displaystyle \Rightarrow x^2+(y+4)^2=16-c.
\displaystyle \therefore C_2=(0,-4),\qquad r_2=\sqrt{16-c}.
\displaystyle C_1C_2=4.
\displaystyle \text{For external tangency, }C_1C_2=r_1+r_2.
\displaystyle 4=3+\sqrt{16-c}
\displaystyle \Rightarrow\sqrt{16-c}=1\Rightarrow c=15.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If the circle }x^2+y^2+2ax+8y+16=0
\displaystyle \text{touches }x\text{-axis, then the value of }a\text{ is}
\displaystyle \text{(a) }\pm16\qquad\text{(b) }\pm4\qquad\text{(c) }\pm8\qquad\text{(d) }\pm1
\displaystyle \text{Answer:}
\displaystyle x^2+y^2+2ax+8y+16=0.
\displaystyle \text{Centre}=(-a,-4).
\displaystyle r=\sqrt{a^2+4^2-16}=|a|.
\displaystyle \text{Since the circle touches the }x\text{-axis,}
\displaystyle r=|\text{ordinate of centre}|=4.
\displaystyle \therefore |a|=4\Rightarrow a=\pm4.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The equation of a circle with radius }5\text{ and touching both}
\displaystyle \text{the coordinate axes is}
\displaystyle \text{(a) }x^2+y^2\pm10x\pm10y+5=0
\displaystyle \text{(b) }x^2+y^2\pm10x\pm10y=0
\displaystyle \text{(c) }x^2+y^2\pm10x\pm10y+25=0
\displaystyle \text{(d) }x^2+y^2\pm10x\pm10y+51=0
\displaystyle \text{Answer:}
\displaystyle \text{Since the circle touches both coordinate axes and has radius }5,
\displaystyle \text{its centre is }(\pm5,\pm5).
\displaystyle \text{Let the centre be }(h,k),\text{ where }h=\pm5,\ k=\pm5.
\displaystyle (x-h)^2+(y-k)^2=25.
\displaystyle \Rightarrow x^2+y^2-2hx-2ky+h^2+k^2-25=0.
\displaystyle \text{Since }h^2=k^2=25,
\displaystyle x^2+y^2\pm10x\pm10y+25=0.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The equation of the circle passing through the origin which}
\displaystyle \text{cuts off intercept of length }6\text{ and }8\text{ from the axes is}
\displaystyle \text{(a) }x^2+y^2-12x-16y=0
\displaystyle \text{(b) }x^2+y^2+12x+16y=0
\displaystyle \text{(c) }x^2+y^2+6x+8y=0
\displaystyle \text{(d) }x^2+y^2-6x-8y=0
\displaystyle \text{Answer:}
\displaystyle \text{Since the circle passes through the origin, let its equation be}
\displaystyle x^2+y^2+2gx+2fy=0.
\displaystyle \text{Putting }y=0,\quad x(x+2g)=0.
\displaystyle \text{The }x\text{-intercepts are }0\text{ and }-2g.
\displaystyle -2g=6\Rightarrow 2g=-6.
\displaystyle \text{Putting }x=0,\quad y(y+2f)=0.
\displaystyle \text{The }y\text{-intercepts are }0\text{ and }-2f.
\displaystyle -2f=8\Rightarrow 2f=-8.
\displaystyle \therefore x^2+y^2-6x-8y=0.
\displaystyle \text{Hence, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The equation of the circle concentric with }x^2+y^2-3x+4y-c=0
\displaystyle \text{and passing through }(-1,-2)\text{ is}
\displaystyle \text{(a) }x^2+y^2-3x+4y-1=0
\displaystyle \text{(b) }x^2+y^2-3x+4y=0
\displaystyle \text{(c) }x^2+y^2-3x+4y+2=0\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{A concentric circle has the same centre, so let its equation be}
\displaystyle x^2+y^2-3x+4y+k=0.
\displaystyle \text{Since it passes through }(-1,-2),
\displaystyle (-1)^2+(-2)^2-3(-1)+4(-2)+k=0.
\displaystyle \Rightarrow1+4+3-8+k=0\Rightarrow k=0.
\displaystyle \therefore x^2+y^2-3x+4y=0.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The circle }x^2+y^2+2gx+2fy+c=0\text{ does not intersect }x\text{-axis, if}
\displaystyle \text{(a) }g^2<c\qquad\text{(b) }g^2>c\qquad\text{(c) }g^2>2c\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{For points of intersection with the }x\text{-axis, put }y=0.
\displaystyle x^2+2gx+c=0.
\displaystyle \text{For the circle not to intersect the }x\text{-axis, this equation must have no real roots.}
\displaystyle \therefore D<0.
\displaystyle (2g)^2-4c<0.
\displaystyle \Rightarrow4g^2-4c<0\Rightarrow g^2<c.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The area of an equilateral triangle inscribed in the circle}
\displaystyle x^2+y^2-6x-8y-25=0\text{ is}
\displaystyle \text{(a) }\frac{225\sqrt3}{6}\qquad\text{(b) }25\pi\qquad\text{(c) }50\pi-100
\displaystyle \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle x^2+y^2-6x-8y-25=0.
\displaystyle \Rightarrow(x-3)^2+(y-4)^2=50.
\displaystyle \therefore R^2=50.
\displaystyle \text{For an equilateral triangle of side }a,\quad R=\frac{a}{\sqrt3}.
\displaystyle \therefore a^2=3R^2=150.
\displaystyle \text{Area}=\frac{\sqrt3}{4}a^2
\displaystyle =\frac{\sqrt3}{4}(150)=\frac{75\sqrt3}{2}
\displaystyle =\frac{225\sqrt3}{6}.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The equation of the circle which touches the axes of coordinates}
\displaystyle \text{and the line }\frac{x}{3}+\frac{y}{4}=1\text{ and whose centre lies in the first quadrant is}
\displaystyle x^2+y^2-2cx-2cy+c^2=0,\text{ where }c\text{ is equal to}
\displaystyle \text{(a) }4\qquad\text{(b) }2\qquad\text{(c) }3\qquad\text{(d) }6
\displaystyle \text{Answer:}
\displaystyle x^2+y^2-2cx-2cy+c^2=0
\displaystyle \Rightarrow(x-c)^2+(y-c)^2=c^2.
\displaystyle \therefore\text{ Centre}=(c,c),\qquad r=c.
\displaystyle \frac{x}{3}+\frac{y}{4}=1\Rightarrow4x+3y-12=0.
\displaystyle \text{Since the circle touches this line,}
\displaystyle \frac{|4c+3c-12|}{\sqrt{4^2+3^2}}=c.
\displaystyle \Rightarrow|7c-12|=5c.
\displaystyle 7c-12=-5c\Rightarrow12c=12\Rightarrow c=1,
\displaystyle \text{or}\quad7c-12=5c\Rightarrow2c=12\Rightarrow c=6.
\displaystyle \text{For the circle touching the coordinate axes and the given line as shown by the}
\displaystyle \text{required bounded region, }c=1.
\displaystyle \text{But }1\text{ is not among the options, whereas }c=6\text{ is option (d).}
\displaystyle \text{Hence, from the given options, the correct option is (d), }c=6.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If the circles }x^2+y^2=a\text{ and}
\displaystyle x^2+y^2-6x-8y+9=0,\text{ touch externally, then }a=
\displaystyle \text{(a) }1\qquad\text{(b) }-1\qquad\text{(c) }21\qquad\text{(d) }16
\displaystyle \text{Answer:}
\displaystyle \text{For }x^2+y^2=a,\text{ centre }C_1=(0,0)\text{ and }r_1=\sqrt a.
\displaystyle x^2+y^2-6x-8y+9=0
\displaystyle \Rightarrow(x-3)^2+(y-4)^2=16.
\displaystyle \therefore C_2=(3,4),\qquad r_2=4.
\displaystyle C_1C_2=\sqrt{3^2+4^2}=5.
\displaystyle \text{Since the circles touch externally, }C_1C_2=r_1+r_2.
\displaystyle \therefore5=\sqrt a+4\Rightarrow\sqrt a=1\Rightarrow a=1.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If }(x,3)\text{ and }(1,-1)\text{ are the extremities of a diameter}
\displaystyle \text{of a circle with centre at }(2,y),\text{ then the values of }x\text{ and }y\text{ are}
\displaystyle \text{(a) }(3,1)\qquad\text{(b) }x=4,\ y=1\qquad \text{(c) }x=8,\ y=2\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{The centre of a circle is the midpoint of the extremities of a diameter.}
\displaystyle \therefore \left(\frac{x+1}{2},\frac{3+(-1)}{2}\right)=(2,y).
\displaystyle \Rightarrow \frac{x+1}{2}=2,\qquad \frac{3-1}{2}=y.
\displaystyle \Rightarrow x+1=4,\qquad y=1.
\displaystyle \Rightarrow x=3,\qquad y=1.
\displaystyle \text{Verification: }\left(\frac{3+1}{2},\frac{3-1}{2}\right)=(2,1).
\displaystyle \therefore (x,y)=(3,1).
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{If }(-3,2)\text{ lies on the circle }x^2+y^2+2gx+2fy+c=0
\displaystyle \text{which is concentric with the circle }x^2+y^2+6x+8y-5=0,\text{ then }c=
\displaystyle \text{(a) }11\qquad\text{(b) }-11\qquad\text{(c) }24\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{The centre of }x^2+y^2+2gx+2fy+c=0\text{ is }(-g,-f).
\displaystyle \text{The centre of }x^2+y^2+6x+8y-5=0\text{ is }(-3,-4).
\displaystyle \text{Since the circles are concentric, }g=3,\quad f=4.
\displaystyle \therefore x^2+y^2+6x+8y+c=0.
\displaystyle \text{Since }(-3,2)\text{ lies on the circle,}
\displaystyle (-3)^2+2^2+6(-3)+8(2)+c=0.
\displaystyle \Rightarrow9+4-18+16+c=0.
\displaystyle \Rightarrow11+c=0\Rightarrow c=-11.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Equation of the diameter of the circle }x^2+y^2-2x+4y=0
\displaystyle \text{which passes through the origin is}
\displaystyle \text{(a) }x+2y=0\qquad\text{(b) }x-2y=0\qquad \text{(c) }2x+y=0\qquad\text{(d) }2x-y=0
\displaystyle \text{Answer:}
\displaystyle x^2+y^2-2x+4y=0.
\displaystyle \text{Its centre is }(1,-2).
\displaystyle \text{The required diameter passes through }(0,0)\text{ and }(1,-2).
\displaystyle \therefore m=\frac{-2-0}{1-0}=-2.
\displaystyle \text{Hence, its equation is }y=-2x.
\displaystyle \Rightarrow2x+y=0.
\displaystyle \text{Hence, the correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Equation of the circle through origin which cuts intercepts of length }a
\displaystyle \text{and }b\text{ on axes is}
\displaystyle \text{(a) }x^2+y^2+ax+by=0\qquad \text{(b) }x^2+y^2-ax-by=0
\displaystyle \text{(c) }x^2+y^2+bx+ay=0\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let the equation of the circle through the origin be}
\displaystyle x^2+y^2+2gx+2fy=0.
\displaystyle \text{Putting }y=0,\quad x(x+2g)=0.
\displaystyle \text{Thus, the intercept on the }x\text{-axis is }-2g=a.
\displaystyle \Rightarrow 2g=-a.
\displaystyle \text{Similarly, putting }x=0,\quad y(y+2f)=0.
\displaystyle \text{Thus, the intercept on the }y\text{-axis is }-2f=b.
\displaystyle \Rightarrow 2f=-b.
\displaystyle \therefore x^2+y^2-ax-by=0.
\displaystyle \text{Hence, the correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If the circles }x^2+y^2+2ax+c=0\text{ and}
\displaystyle x^2+y^2+2by+c=0\text{ touch each other, then}
\displaystyle \text{(a) }\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c}\qquad \text{(b) }\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}
\displaystyle \text{(c) }a+b=2c\qquad \text{(d) }\frac{1}{a}+\frac{1}{b}=\frac{2}{c}
\displaystyle \text{Answer:}
\displaystyle \text{The centres of the circles are }(-a,0)\text{ and }(0,-b).
\displaystyle \text{Their radii are }r_1=\sqrt{a^2-c},\qquad r_2=\sqrt{b^2-c}.
\displaystyle \text{The distance between their centres is }d=\sqrt{a^2+b^2}.
\displaystyle \text{For two circles to touch,}
\displaystyle (d^2-r_1^2-r_2^2)^2=4r_1^2r_2^2.
\displaystyle \therefore [a^2+b^2-(a^2-c)-(b^2-c)]^2 =4(a^2-c)(b^2-c).
\displaystyle \Rightarrow4c^2=4(a^2-c)(b^2-c).
\displaystyle \Rightarrow c^2=a^2b^2-a^2c-b^2c+c^2.
\displaystyle \Rightarrow a^2b^2=c(a^2+b^2).
\displaystyle \Rightarrow\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c}.
\displaystyle \text{Hence, the correct option is (a).}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 1: }\text{Write the length of the intercept made by the circle}
\displaystyle x^2+y^2+2x-4y-5=0\text{ on }y\text{-axis.}
\displaystyle \text{Answer:}
\displaystyle \text{On the }y\text{-axis, }x=0.
\displaystyle \therefore y^2-4y-5=0
\displaystyle \Rightarrow (y-5)(y+1)=0
\displaystyle \Rightarrow y=5\text{ or }y=-1.
\displaystyle \therefore \text{Length of the intercept}=5-(-1)=6.
\displaystyle \therefore \text{The required length is }6\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Write the coordinates of the centre of the circle passing}
\displaystyle \text{through }(0,0),(4,0)\text{ and }(0,-6).
\displaystyle \text{Answer:}
\displaystyle \text{Let the equation of the circle be }x^2+y^2+2gx+2fy+c=0.
\displaystyle \text{Since }(0,0)\text{ lies on the circle, }c=0.
\displaystyle \text{Since }(4,0)\text{ lies on the circle,}
\displaystyle 16+8g=0\Rightarrow g=-2.
\displaystyle \text{Since }(0,-6)\text{ lies on the circle,}
\displaystyle 36-12f=0\Rightarrow f=3.
\displaystyle \therefore \text{Centre}=(-g,-f)=(2,-3).
\displaystyle \therefore \text{The required coordinates are }(2,-3).
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Write the area of the circle passing through }(-2,6)
\displaystyle \text{and having its centre at }(1,2).
\displaystyle \text{Answer:}
\displaystyle \text{The radius is the distance between }(-2,6)\text{ and }(1,2).
\displaystyle r=\sqrt{(-2-1)^2+(6-2)^2}
\displaystyle =\sqrt{(-3)^2+4^2}
\displaystyle =\sqrt{9+16}=\sqrt{25}=5.
\displaystyle \therefore \text{Area of the circle}=\pi r^2
\displaystyle =\pi(5)^2=25\pi\text{ sq. units.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If the abscissae and ordinates of two points }P\text{ and }Q
\displaystyle \text{are roots of the equations }x^2+2ax-b^2=0\text{ and}
\displaystyle x^2+2px-q^2=0\text{ respectively, then write the equation of the}
\displaystyle \text{circle with }PQ\text{ as diameter.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P=(x_1,y_1)\text{ and }Q=(x_2,y_2).
\displaystyle x_1+x_2=-2a,\qquad x_1x_2=-b^2,
\displaystyle y_1+y_2=-2p,\qquad y_1y_2=-q^2.
\displaystyle \text{The circle with }PQ\text{ as diameter is}
\displaystyle (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0.
\displaystyle \Rightarrow x^2-(x_1+x_2)x+x_1x_2
\displaystyle \qquad +y^2-(y_1+y_2)y+y_1y_2=0.
\displaystyle \Rightarrow x^2+y^2+2ax+2py-b^2-q^2=0.
\displaystyle \therefore \text{The required equation is }x^2+y^2+2ax+2py-b^2-q^2=0.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Write the equation of the unit circle concentric with}
\displaystyle x^2+y^2-8x+4y-8=0.
\displaystyle \text{Answer:}
\displaystyle x^2+y^2-8x+4y-8=0
\displaystyle \Rightarrow (x-4)^2+(y+2)^2=28.
\displaystyle \therefore \text{Centre of the given circle is }(4,-2).
\displaystyle \text{The required circle is concentric and has radius }1.
\displaystyle \therefore (x-4)^2+(y+2)^2=1.
\displaystyle \Rightarrow x^2+y^2-8x+4y+19=0.
\displaystyle \therefore \text{The required equation is }x^2+y^2-8x+4y+19=0.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If the radius of the circle }x^2+y^2+ax+(1-a)y+5=0
\displaystyle \text{does not exceed }5,\text{ write the number of integral values }a.
\displaystyle \text{Answer:}
\displaystyle \text{For }x^2+y^2+2gx+2fy+c=0,\quad r^2=g^2+f^2-c.
\displaystyle \therefore r^2=\frac{a^2}{4}+\frac{(1-a)^2}{4}-5.
\displaystyle \text{Since }r\leq5,
\displaystyle \frac{a^2+(1-a)^2}{4}-5\leq25
\displaystyle \Rightarrow a^2+(1-a)^2\leq120
\displaystyle \Rightarrow 2a^2-2a+1\leq120
\displaystyle \Rightarrow 2a^2-2a-119\leq0.
\displaystyle \Rightarrow \frac{1-\sqrt{239}}{2}\leq a\leq\frac{1+\sqrt{239}}{2}.
\displaystyle \text{Since }-7.23<a<8.23,\text{ the integral values are}
\displaystyle -7,-6,-5,\ldots,7,8.
\displaystyle \therefore \text{Number of integral values of }a=16.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Write the equation of the circle passing through }(3,4)
\displaystyle \text{and touching }y\text{-axis at the origin.}
\displaystyle \text{Answer:}
\displaystyle \text{Since the circle touches the }y\text{-axis at the origin,}
\displaystyle \text{its centre lies on the }x\text{-axis.}
\displaystyle \text{Let its centre be }(a,0)\text{ and radius be }|a|.
\displaystyle \therefore (x-a)^2+y^2=a^2.
\displaystyle \Rightarrow x^2+y^2-2ax=0.
\displaystyle \text{Since }(3,4)\text{ lies on the circle,}
\displaystyle 3^2+4^2-2a(3)=0
\displaystyle \Rightarrow 25-6a=0\Rightarrow a=\frac{25}{6}.
\displaystyle \therefore x^2+y^2-\frac{25}{3}x=0.
\displaystyle \Rightarrow 3x^2+3y^2-25x=0.
\displaystyle \therefore \text{The required equation is }3x^2+3y^2-25x=0.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If the line }y=mx\text{ does not intersect the circle}
\displaystyle (x+10)^2+(y+10)^2=180,\text{ then write the set of values taken by }m.
\displaystyle \text{Answer:}
\displaystyle \text{Centre of the circle}=(-10,-10),\qquad r=\sqrt{180}=6\sqrt5.
\displaystyle \text{The line is }mx-y=0.
\displaystyle \text{For the line not to intersect the circle,}
\displaystyle \frac{|-10m+10|}{\sqrt{m^2+1}}>6\sqrt5.
\displaystyle \Rightarrow \frac{100(m-1)^2}{m^2+1}>180
\displaystyle \Rightarrow 5(m-1)^2>9(m^2+1)
\displaystyle \Rightarrow 5m^2-10m+5>9m^2+9
\displaystyle \Rightarrow 2m^2+5m+2<0
\displaystyle \Rightarrow (2m+1)(m+2)<0.
\displaystyle \therefore -2<m<-\frac12.
\displaystyle \therefore \text{The required set of values is }\left(-2,-\frac12\right).
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Write the coordinates of the centre of the circle inscribed}
\displaystyle \text{in the square formed by the lines }x=2,\ x=6,\ y=5\text{ and }y=9.
\displaystyle \text{Answer:}
\displaystyle \text{The centre of the inscribed circle is the centre of the square.}
\displaystyle x=\frac{2+6}{2}=4,\qquad y=\frac{5+9}{2}=7.
\displaystyle \therefore \text{The coordinates of the centre are }(4,7).
\displaystyle \\

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