Class 11: Circles – Multiple Choice Questions

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

$\displaystyle \textbf{Question 2. }\text{If }2x^2+\lambda xy+2y^2+(\lambda-4)x+6y-5=0\text{ is the equation of a circle,} \\ \text{then its radius is}$
$\displaystyle \text{(a) }3\sqrt2\qquad \text{(b) }2\sqrt3\qquad \text{(c) }2\sqrt2\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the equation to represent a circle, coefficient of }xy\text{ must be }0$
$\displaystyle \therefore \lambda=0$
$\displaystyle \text{Equation becomes }2x^2+2y^2-4x+6y-5=0$
$\displaystyle x^2+y^2-2x+3y-\frac{5}{2}=0$
$\displaystyle x^2+y^2+2gx+2fy+c=0$
$\displaystyle g=-1,\quad f=\frac{3}{2},\quad c=-\frac{5}{2}$
$\displaystyle r=\sqrt{g^2+f^2-c}$
$\displaystyle =\sqrt{1+\frac{9}{4}+\frac{5}{2}}$
$\displaystyle =\sqrt{\frac{23}{4}}$
$\displaystyle =\frac{\sqrt{23}}{2}$
$\displaystyle \therefore \text{Correct option is (d)}$
$\\$

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

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

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

$\displaystyle \textbf{Question 6. }\text{The number of integral values of }\lambda\text{ for which the equation }$
$\displaystyle x^2+y^2+\lambda x+(1-\lambda)y+5=0 \text{is the equation of a circle whose radius cannot exceed }5,\text{ is}$
$\displaystyle \text{(a) }14\qquad \text{(b) }18\qquad \text{(c) }16\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle x^2+y^2+\lambda x+(1-\lambda)y+5=0$
$\displaystyle 2g=\lambda,\quad 2f=1-\lambda,\quad c=5$
$\displaystyle r^2=g^2+f^2-c$
$\displaystyle =\frac{\lambda^2}{4}+\frac{(1-\lambda)^2}{4}-5$
$\displaystyle =\frac{2\lambda^2-2\lambda+1}{4}-5$
$\displaystyle =\frac{2\lambda^2-2\lambda-19}{4}$
$\displaystyle \text{Since radius cannot exceed }5,$
$\displaystyle r^2\leq25$
$\displaystyle \frac{2\lambda^2-2\lambda-19}{4}\leq25$
$\displaystyle 2\lambda^2-2\lambda-119\leq0$
$\displaystyle \lambda=\frac{2\pm\sqrt{956}}{4}$
$\displaystyle \lambda=\frac{1\pm\sqrt{239}}{2}$
$\displaystyle -7.23\leq\lambda\leq8.23$
$\displaystyle \therefore \text{Integral values are }-7,-6,\ldots,8$
$\displaystyle \text{Number of integral values}=16$
$\displaystyle \therefore \text{Correct option is (c)}$
$\\$

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

$\displaystyle \textbf{Question 8. }\text{If the centroid of an equilateral triangle is }(1,1)\text{ and its one vertex is }(-1,2),$
$\displaystyle \text{then the equation of its circumcircle is}$
$\displaystyle \text{(a) }x^2+y^2-2x-2y-3=0\qquad \text{(b) }x^2+y^2+2x-2y-3=0$
$\displaystyle \text{(c) }x^2+y^2+2x+2y-3=0\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In an equilateral triangle, centroid and circumcentre coincide.}$
$\displaystyle \therefore \text{Centre of circumcircle}=(1,1)$
$\displaystyle \text{Radius}=\sqrt{(-1-1)^2+(2-1)^2}$
$\displaystyle =\sqrt{4+1}=\sqrt5$
$\displaystyle \therefore (x-1)^2+(y-1)^2=5$
$\displaystyle x^2+y^2-2x-2y-3=0$
$\displaystyle \therefore \text{Correct option is (a)}$
$\\$

$\displaystyle \textbf{Question 9. }\text{If the point }(2,k)\text{ lies outside the circles } \\ x^2+y^2+x-2y-14=0\text{ and }x^2+y^2=13,$
$\displaystyle \text{then }k\text{ lies in the interval}$
$\displaystyle \text{(a) }(-3,-2)\cup(3,4)\qquad \text{(b) }(-3,4)$
$\displaystyle \text{(c) }(-\infty,-3)\cup(4,\infty)\qquad \text{(d) }(-\infty,-2)\cup(3,\infty)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }x^2+y^2+x-2y-14=0,$
$\displaystyle x^2+y^2+x-2y=14$
$\displaystyle \text{Substituting }(2,k),$
$\displaystyle 4+k^2+2-2k>14$
$\displaystyle k^2-2k-8>0$
$\displaystyle (k-4)(k+2)>0$
$\displaystyle k<-2\text{ or }k>4$
$\displaystyle \text{For }x^2+y^2=13,$
$\displaystyle 4+k^2>13$
$\displaystyle k^2>9$
$\displaystyle k<-3\text{ or }k>3$
$\displaystyle \text{Combining both conditions,}$
$\displaystyle k\in(-\infty,-3)\cup(4,\infty)$
$\displaystyle \therefore \text{Correct option is (c)}$
$\\$

$\displaystyle \textbf{Question 10. }\text{If the point }(\lambda,\lambda+1)\text{ lies inside the region bounded by the curve }$
$\displaystyle x=\sqrt{25-y^2} \text{ and }y\text{-axis, then }\lambda\text{ belongs to the interval}$
$\displaystyle \text{(a) }(-1,3)\qquad \text{(b) }(-4,3)\qquad \text{(c) }(-\infty,-4)\cup(3,\infty)\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle x=\sqrt{25-y^2}$
$\displaystyle \Rightarrow x^2+y^2=25,\quad x\geq0$
$\displaystyle \text{Inside the bounded region implies}$
$\displaystyle x^2+y^2<25\quad \text{and}\quad x>0$
$\displaystyle \lambda^2+(\lambda+1)^2<25$
$\displaystyle 2\lambda^2+2\lambda+1<25$
$\displaystyle \lambda^2+\lambda-12<0$
$\displaystyle (\lambda+4)(\lambda-3)<0$
$\displaystyle -4<\lambda<3$
$\displaystyle \therefore \text{Correct option is (b)}$
$\\$

$\displaystyle \textbf{Question 11. }\text{The equation of the incircle formed by the coordinate axes and the line } \\ 4x+3y=6\text{ is}$
$\displaystyle \text{(a) }x^2+y^2-6x-6y+9=0\qquad \text{(b) }4(x^2+y^2-x-y)+1=0$
$\displaystyle \text{(c) }4(x^2+y^2+x+y)+1=0\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The triangle is formed by }x=0,\ y=0\text{ and }4x+3y=6$
$\displaystyle \text{Let the incenter be }(r,r)$
$\displaystyle \text{Distance from }(r,r)\text{ to }4x+3y-6=0\text{ equals }r$
$\displaystyle \frac{|4r+3r-6|}{\sqrt{4^2+3^2}}=r$
$\displaystyle \frac{6-7r}{5}=r$
$\displaystyle 6=12r$
$\displaystyle r=\frac{1}{2}$
$\displaystyle \therefore \text{Equation of incircle is}$
$\displaystyle \left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=\frac{1}{4}$
$\displaystyle x^2+y^2-x-y+\frac{1}{4}=0$
$\displaystyle \therefore 4(x^2+y^2-x-y)+1=0$
$\displaystyle \therefore \text{Correct option is (b)}$
$\\$

$\displaystyle \textbf{Question 12. }\text{If the circles }x^2+y^2=9\text{ and }x^2+y^2+8y+c=0\text{ touch each other, then } \\ c\text{ is equal to}$
$\displaystyle \text{(a) }15\qquad \text{(b) }-15\qquad \text{(c) }16\qquad \text{(d) }-16$
$\displaystyle \text{Answer:}$
$\displaystyle x^2+y^2=9$
$\displaystyle \text{Centre }C_1=(0,0),\quad r_1=3$
$\displaystyle x^2+y^2+8y+c=0$
$\displaystyle x^2+(y+4)^2=16-c$
$\displaystyle \text{Centre }C_2=(0,-4),\quad r_2=\sqrt{16-c}$
$\displaystyle \text{Distance between centres}=4$
$\displaystyle \text{For touching externally, }r_1+r_2=4$
$\displaystyle 3+\sqrt{16-c}=4$
$\displaystyle \sqrt{16-c}=1$
$\displaystyle c=15$
$\displaystyle \therefore \text{Correct option is (a)}$
$\\$

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

$\displaystyle \textbf{Question 14. }\text{The equation of a circle with radius }5\text{ and touching both the coordinate axes is}$
$\displaystyle \text{(a) }x^2+y^2\pm10x\pm10y+5=0\qquad \text{(b) }x^2+y^2\pm10x\pm10y=0$
$\displaystyle \text{(c) }x^2+y^2\pm10x\pm10y+25=0\qquad \text{(d) }x^2+y^2\pm10x\pm10y+51=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since the circle touches both coordinate axes and radius is }5,$
$\displaystyle \text{its centre is }(\pm5,\pm5)$
$\displaystyle \text{Equation is }(x\mp5)^2+(y\mp5)^2=25$
$\displaystyle x^2+y^2\mp10x\mp10y+25=0$
$\displaystyle \therefore \text{Correct option is (c)}$
$\\$

$\displaystyle \textbf{Question 15. }\text{The equation of the circle passing through the origin which cuts off} \\ \text{intercept of length }6\text{ and }8\text{ from the axes is}$
$\displaystyle \text{(a) }x^2+y^2-12x-16y=0\qquad \text{(b) }x^2+y^2+12x+16y=0$
$\displaystyle \text{(c) }x^2+y^2+6x+8y=0\qquad \text{(d) }x^2+y^2-6x-8y=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Circle passes through origin, so its equation is}$
$\displaystyle x^2+y^2+2gx+2fy=0$
$\displaystyle \text{Putting }y=0,\quad x^2+2gx=0$
$\displaystyle x(x+2g)=0$
$\displaystyle \text{Length of intercept on }x\text{-axis}=|2g|=6$
$\displaystyle \therefore 2g=\pm6$
$\displaystyle \text{Putting }x=0,\quad y^2+2fy=0$
$\displaystyle y(y+2f)=0$
$\displaystyle \text{Length of intercept on }y\text{-axis}=|2f|=8$
$\displaystyle \therefore 2f=\pm8$
$\displaystyle \text{Hence one possible equation is }x^2+y^2-6x-8y=0$
$\displaystyle \therefore \text{Correct option is (d)}$
$\\$

$\displaystyle \textbf{Question 16. }\text{The equation of the circle concentric with } \\ x^2+y^2-3x+4y-c=0\text{ and passing through }(-1,-2)\text{ is}$
$\displaystyle \text{(a) }x^2+y^2-3x+4y-1=0\qquad \text{(b) }x^2+y^2-3x+4y=0$
$\displaystyle \text{(c) }x^2+y^2-3x+4y+2=0\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{A concentric circle will have equation }x^2+y^2-3x+4y+k=0$
$\displaystyle \text{Since it passes through }(-1,-2),$
$\displaystyle 1+4+3-8+k=0$
$\displaystyle k=0$
$\displaystyle \therefore x^2+y^2-3x+4y=0$
$\displaystyle \therefore \text{Correct option is (b)}$
$\\$

$\displaystyle \textbf{Question 17. }\text{The circle }x^2+y^2+2gx+2fy+c=0\text{ does not intersect }x\text{-axis, if}$
$\displaystyle \text{(a) }g^2<c\qquad \text{(b) }g^2>c\qquad \text{(c) }g^2>2c\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For intersection with }x\text{-axis, put }y=0$
$\displaystyle x^2+2gx+c=0$
$\displaystyle \text{For no intersection with }x\text{-axis, this quadratic has no real roots.}$
$\displaystyle D<0$
$\displaystyle (2g)^2-4c<0$
$\displaystyle 4g^2-4c<0$
$\displaystyle g^2<c$
$\displaystyle \therefore \text{Correct option is (a)}$
$\\$

$\displaystyle \textbf{Question 18. }\text{The area of an equilateral triangle inscribed in the circle } \\ x^2+y^2-6x-8y-25=0\text{ is}$
$\displaystyle \text{(a) }\frac{225\sqrt3}{6}\qquad \text{(b) }25\pi\qquad \text{(c) }50\pi-100\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle x^2+y^2-6x-8y-25=0$
$\displaystyle (x-3)^2+(y-4)^2=50$
$\displaystyle \therefore R=5\sqrt2$
$\displaystyle \text{For an equilateral triangle, circumradius }R=\frac{s}{\sqrt3}$
$\displaystyle \therefore s=R\sqrt3=5\sqrt6$
$\displaystyle \text{Area}=\frac{\sqrt3}{4}s^2$
$\displaystyle =\frac{\sqrt3}{4}(150)$
$\displaystyle =\frac{75\sqrt3}{2}$
$\displaystyle =\frac{225\sqrt3}{6}$
$\displaystyle \therefore \text{Correct option is (a)}$
$\\$

$\displaystyle \textbf{Question 19. }\text{The equation of the circle which touches the axes of coordinates} \\ \text{and the line } \frac{x}{3}+\frac{y}{4}=1 \text{and whose centre lies in the first quadrant is } \\ x^2+y^2-2cx-2cy+c^2=0,\text{ where }c\text{ is equal to}$
$\displaystyle \text{(a) }4\qquad \text{(b) }2\qquad \text{(c) }3\qquad \text{(d) }6$
$\displaystyle \text{Answer:}$
$\displaystyle x^2+y^2-2cx-2cy+c^2=0$
$\displaystyle \Rightarrow (x-c)^2+(y-c)^2=c^2$
$\displaystyle \therefore \text{Centre}=(c,c)\text{ and radius}=c$
$\displaystyle \frac{x}{3}+\frac{y}{4}=1$
$\displaystyle \Rightarrow 4x+3y-12=0$
$\displaystyle \text{Since the circle touches this line,}$
$\displaystyle \frac{|4c+3c-12|}{\sqrt{4^2+3^2}}=c$
$\displaystyle \frac{|7c-12|}{5}=c$
$\displaystyle |7c-12|=5c$
$\displaystyle 7c-12=5c\quad \text{or}\quad 7c-12=-5c$
$\displaystyle c=6\quad \text{or}\quad c=1$
$\displaystyle \text{From the given options, }c=6$
$\displaystyle \therefore \text{Correct option is (d)}$
$\\$

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

$\displaystyle \textbf{Question 21. }\text{If }(x,3)\text{ and }(3,5)\text{ are the extremities of a diameter of a circle with centre at }(2,y),$
$\displaystyle \text{then the values of }x\text{ and }y\text{ are}$
$\displaystyle \text{(a) }(3,1)\qquad \text{(b) }x=4,\ y=1\qquad \text{(c) }x=8,\ y=2\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Centre is the midpoint of the endpoints of the diameter.}$
$\displaystyle \left(\frac{x+3}{2},\frac{3+5}{2}\right)=(2,y)$
$\displaystyle \frac{x+3}{2}=2,\quad y=4$
$\displaystyle x+3=4$
$\displaystyle x=1,\quad y=4$
$\displaystyle \therefore \text{Correct option is (d)}$
$\\$

$\displaystyle \textbf{Question 22. }\text{If }(-3,2)\text{ lies on the circle }x^2+y^2+2gx+2fy+c=0$
$\displaystyle \text{which is concentric with the circle} x^2+y^2+6x+8y-5=0,\text{ then }c=$
$\displaystyle \text{(a) }11\qquad \text{(b) }-11\qquad \text{(c) }24\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle x^2+y^2+6x+8y-5=0$
$\displaystyle \text{Centre}=(-3,-4)$
$\displaystyle \therefore -g=-3,\quad -f=-4$
$\displaystyle g=3,\quad f=4$
$\displaystyle \text{Required circle is }x^2+y^2+6x+8y+c=0$
$\displaystyle \text{Since }(-3,2)\text{ lies on it,}$
$\displaystyle 9+4+6(-3)+8(2)+c=0$
$\displaystyle 13-18+16+c=0$
$\displaystyle c=-11$
$\displaystyle \therefore \text{Correct option is (b)}$
$\\$

$\displaystyle \textbf{Question 23. }\text{Equation of the diameter of the circle } \\ x^2+y^2-2x+4y=0\text{ which passes through the origin is}$
$\displaystyle \text{(a) }x+2y=0\qquad \text{(b) }x-2y=0\qquad \text{(c) }2x+y=0\qquad \text{(d) }2x-y=0$
$\displaystyle \text{Answer:}$
$\displaystyle x^2+y^2-2x+4y=0$
$\displaystyle x^2+y^2+2gx+2fy=0$
$\displaystyle 2g=-2,\quad 2f=4$
$\displaystyle g=-1,\quad f=2$
$\displaystyle \therefore \text{Centre}=(-g,-f)=(1,-2)$
$\displaystyle \text{Diameter through origin passes through }(0,0)\text{ and }(1,-2)$
$\displaystyle \text{Equation of required diameter is }y=-2x$
$\displaystyle \therefore 2x+y=0$
$\displaystyle \therefore \text{Correct option is (c)}$
$\\$

$\displaystyle \textbf{Question 24. }\text{Equation of the circle through origin which cuts intercepts of length } \\ a\text{ and }b\text{ on axes is}$
$\displaystyle \text{(a) }x^2+y^2+ax+by=0\qquad \text{(b) }x^2+y^2-ax-by=0$
$\displaystyle \text{(c) }x^2+y^2+bx+ay=0\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since the circle passes through origin, its equation is}$
$\displaystyle x^2+y^2+2gx+2fy=0$
$\displaystyle \text{Putting }y=0,$
$\displaystyle x^2+2gx=0$
$\displaystyle x(x+2g)=0$
$\displaystyle \text{Length of intercept on }x\text{-axis}=|2g|=a$
$\displaystyle \text{Putting }x=0,$
$\displaystyle y^2+2fy=0$
$\displaystyle y(y+2f)=0$
$\displaystyle \text{Length of intercept on }y\text{-axis}=|2f|=b$
$\displaystyle \text{Hence one possible equation is }x^2+y^2-ax-by=0$
$\displaystyle \therefore \text{Correct option is (b)}$
$\\$

$\displaystyle \textbf{Question 25. }\text{If the circles }x^2+y^2+2ax+c=0\text{ and } \\ x^2+y^2+2by+c=0\text{ touch each other, then}$
$\displaystyle \text{(a) }\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c}\qquad \text{(b) }\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$
$\displaystyle \text{(c) }a+b=2c\qquad \text{(d) }\frac{1}{a}+\frac{1}{b}=\frac{2}{c}$
$\displaystyle \text{Answer:}$
$\displaystyle C_1=(-a,0),\quad r_1=\sqrt{a^2-c}$
$\displaystyle C_2=(0,-b),\quad r_2=\sqrt{b^2-c}$
$\displaystyle C_1C_2=\sqrt{a^2+b^2}$
$\displaystyle \text{For the two circles to touch externally,}$
$\displaystyle \sqrt{a^2+b^2}=\sqrt{a^2-c}+\sqrt{b^2-c}$
$\displaystyle \text{Squaring,}$
$\displaystyle a^2+b^2=a^2+b^2-2c+2\sqrt{(a^2-c)(b^2-c)}$
$\displaystyle c=\sqrt{(a^2-c)(b^2-c)}$
$\displaystyle c^2=(a^2-c)(b^2-c)$
$\displaystyle c^2=a^2b^2-a^2c-b^2c+c^2$
$\displaystyle a^2b^2=c(a^2+b^2)$
$\displaystyle \therefore \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c}$
$\displaystyle \therefore \text{Correct option is (a)}$
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