# Inversion Transformation and Inversion Circle

In the figure, O is centre of circle, OP = 4 units, OQ = 8 units and OP' = 16 units, write the relation between P and P'.

Given,
OP = 4 units, OQ = 8 units, OP' = 16 units

Now, We have, OP × OP' = OQ2 or, 4 × 16 = 82
so. 64 = 64

It is like OP × OP' = r2. So P' is inversion point of P and vice-versa with respect to the circle.

Find the inverse image of the point (4, 5) with respect to the circle x2 + y2 = 10

Here,
Centre of the circle = (0, 0)
Radius of the circle (r) = 10 units
Object point P(x, y) = (4, 5)
Inversion point P'(x', y') = ?

We know that,
(x', y') = $(\frac{r^{2}x}{x^{2} \, + \, y^{2}}, \frac{r^{2}y}{x^2 \, + \, y^2})$ = $(\frac{(10)^{2} \times 4}{4^{2} \, + \, 5^{2}}, \frac{(10)^{2} \times 5}{4^2 \, + \, 5^2})$
= $(\frac{400}{41}, \frac{500}{41})$

Hence, the inverse of the point (4, 5) with respect to given circle is (x', y') = $(\frac{400}{41}, \frac{500}{41})$

Find the inverse image of the point (3, 4) about the circle (x - 2)2 + (y - 2)2 = 36.

Here,
Centre of circle (h, k) = (2, 2)
Radius of the circle (r) = 6 units Object point (x, y) = (3, 4)
Inversion point (x', y') = ?

Now, we know,
(x', y') = $(\frac{r^{2}(x - h)}{(x - h)^{2} \, + \, (y - k)^{2}} + h, \frac{r^{2}(y - k)}{(x - h)^{2} \, + \, (y - k)^{2}} + k)$ = $(\frac{36(3 - 2)}{(3 - 2)^{2} \, + \, (4 - 2)^{2}} + 2, \frac{36(4 - 2)}{(3 - 2)^{2} \, + \, (4 - 2)^{2}} + 2)$
= $(\frac{36}{5} + 2, \frac{72}{5} + 2)$
= $(\frac{46}{5}, \frac{82}{5})$

Hence, $(\frac{46}{5}, \frac{82}{5})$ is the required image of the point (3, 4) about the given circle.