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

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

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

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

\displaystyle \textbf{Question 5. }\text{Write the equation of the unit circle concentric with } \\ x^2+y^2-8x+4y-8=0.
\displaystyle \text{Answer:}
\displaystyle x^2+y^2-8x+4y-8=0
\displaystyle \text{Centre}=(-g,-f)=(4,-2)
\displaystyle \text{Unit circle has radius }1
\displaystyle \therefore \text{Required circle is }(x-4)^2+(y+2)^2=1
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\displaystyle \textbf{Question 6. }\text{If the radius of the circle }x^2+y^2+ax+(1-a)y+5=0\text{ does not exceed }5, \\ \text{ write the number of integral values of }a.
\displaystyle \text{Answer:}
\displaystyle \text{For }x^2+y^2+2gx+2fy+c=0,\quad r^2=g^2+f^2-c
\displaystyle 2g=a\Rightarrow g=\frac{a}{2},\quad 2f=1-a\Rightarrow f=\frac{1-a}{2},\quad c=5
\displaystyle r^2=\left(\frac{a}{2}\right)^2+\left(\frac{1-a}{2}\right)^2-5
\displaystyle =\frac{a^2+(1-a)^2}{4}-5
\displaystyle \text{Since radius does not exceed }5,\quad r^2\leq25
\displaystyle \frac{a^2+(1-a)^2}{4}-5\leq25
\displaystyle a^2+(1-a)^2\leq120
\displaystyle 2a^2-2a+1\leq120
\displaystyle 2a^2-2a-119\leq0
\displaystyle \frac{1-\sqrt{239}}{2}\leq a\leq\frac{1+\sqrt{239}}{2}
\displaystyle \therefore -7.23\leq a\leq8.23
\displaystyle \text{Integral values of }a\text{ are }-7,-6,\ldots,8
\displaystyle \therefore \text{Number of integral values}=16
\\

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

\displaystyle \textbf{Question 8. }\text{If the line }y=mx\text{ does not intersect the circle } \\ (x+10)^2+(y+10)^2=180,\text{ then write the set of values taken by }m.
\displaystyle \text{Answer:}
\displaystyle \text{Centre of circle}=(-10,-10),\quad r=\sqrt{180}=6\sqrt5
\displaystyle \text{Line }y=mx\Rightarrow mx-y=0
\displaystyle \text{Distance of centre from line}=\frac{|m(-10)-(-10)|}{\sqrt{m^2+1}}
\displaystyle =\frac{10|1-m|}{\sqrt{m^2+1}}
\displaystyle \text{For no intersection, distance }>\text{ radius}
\displaystyle \frac{10|1-m|}{\sqrt{m^2+1}}>6\sqrt5
\displaystyle 100(1-m)^2>180(m^2+1)
\displaystyle 5(1-2m+m^2)>9(m^2+1)
\displaystyle 5-10m+5m^2>9m^2+9
\displaystyle 4m^2+10m+4<0
\displaystyle 2m^2+5m+2<0
\displaystyle (2m+1)(m+2)<0
\displaystyle \therefore -2<m<-\frac{1}{2}
\\

\displaystyle \textbf{Question 9. }\text{Write the coordinates of the centre of the circle inscribed in the} \\ \text{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\text{-coordinate}=\frac{2+6}{2}=4
\displaystyle y\text{-coordinate}=\frac{5+9}{2}=7
\displaystyle \therefore \text{Centre}=(4,7)
\\

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