### Circle Question 2

#### Question 2 - 25 January - Shift 1

The points of intersection of the line $a x+by =0$, $(a \neq b)$ and the circle $x^{2}+y^{2}-2 x=0$ are $A(\alpha, 0)$ and $B(1, \beta)$. The image of the circle with $AB$ as a diameter in the line $x+y+2=0$ is :

(1) $x^{2}+y^{2}+5 x+5 y+12=0$

(2) $x^{2}+y^{2}+3 x+5 y+8=0$

(3) $x^{2}+y^{2}+3 x+3 y+4=0$

(4) $x^{2}+y^{2}-5 x-5 y+12=0$

## Show Answer

#### Answer: (1)

#### Solution:

#### Formula: Mid Point Formula, Reflection of a point about a line

As $A$ and $B$ satisfy both line and circle we have $\alpha=0 \Rightarrow A(0,0)$ and $\beta=1$ i.e. $B(1,1)$

Centre of circle as $A B$ diameter is $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $=\frac{1}{\sqrt{2}}$

$\therefore$ For image of $\left(\frac{1}{2} ; \frac{1}{2}\right)$ in $x+y+2$ we get $\frac{x-\frac{1}{2}}{1}=\frac{y-\frac{1}{2}}{1}=\frac{-2(3)}{2}$

$ \Rightarrow \text { Image }\left(-\frac{5}{2},-\frac{5}{2}\right) $

$\therefore$ Equation of required circle

$ \left(x+\frac{5}{2}\right)^2+\left(y+\frac{5}{2}\right)^2=\frac{1}{2} $

$ \Rightarrow x^2+y^2+5 x+5 y+\frac{50}{4}-\frac{1}{2}=0 $

$ \Rightarrow x^2+y^2+5 x+5 y+12=0 $