The points of intersection of the line ax+by=0 (a b) and the circle x^2+y^2-2x=0 are A(,0)…

Question

The points of intersection of the line $ax+by=0$ ($a
e b$) and the circle $x^2+y^2-2x=0$ 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:

Accepted Answer

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 $\left(\frac{1}{2}, \frac{1}{2}ight)$ and radius $=\frac{1}{\sqrt{2}}$

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

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

$	herefore $ Equation of required circle

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

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

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

Solutions · 1