Find the image of the circle $x^{2} + y^{2} - 2 x + 4 y - 4 = 0$ in the line $2 x - 3 y + 5 = 0$ .

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Solution

Equation of circle is $x^{2} + y^{2} - 2 x + 4 y - 4 = 0$
Centre $= (- (\frac{- 2}{2}) - (\frac{1}{2}))$
$= (1, - 2)$
Reflection of $(1, - 2)$ over the line $2 x - 3 y + 5 = 0$
is given by $\frac{h - x_{1}}{2} = \frac{k - y_{1}}{- 3} = \frac{- 2 (2 x - 3 y + 5)}{2^{2} + (- 3)^{2}}$
$\frac{h - 1}{2} = \frac{k + 2}{- 3} = \frac{- 2 (2 + 6 + 5)}{13} = - 2$
$h - 1 = - 4$ $k + 2 = 6$
$h = 1 - 4 = - 3$ $k = 4$
Required centre $(- 3, 4)$
Radius of the given circle $= \sqrt{1^{2} + (- 2)^{2} + 4}$
$= \sqrt{1 + 4 + 4}$
$= \sqrt{9}$
$= 3$
Hence the required equation is given by $(x + 3)^{2} + (y - 4)^{2} = 3^{2}$
(or) $x^{2} + y^{2} + 6 x - 8 y + 16 = 0$ .

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