Find the pole of the straight line $9 x + y - 28 = 0$ with respect to the circle $2 x^{2} + 2 y^{2} - 3 x + 5 y - 7 = 0$ .

Open in App

Solution

Let $(h, k)$ be the requited pole. Then its polar with respect to the circle $2 x^{2} + 2 y^{2} - 3 x + 5 y - 7 = 0$ is $2 x h + 2 y k - (3 / 2) (x + h) + (5 / 2) (y + k) - 7 = 0$
or $(4 h - 3) x - (4 k + 5) y - 3 h 5 k - 14 = 0 . . . . . (1)$
Comparing $(1)$ with $9 x + y - 28 = 0$ , we get
$\frac{4 h - 3}{9} = \frac{4 k + 5}{1} = \frac{- 3 h + 5 k - 14}{- 28}$
From first two ratios, we get
$h - 9 k - 12 = 0 . . . . . (2)$
and second and third ratios give
$h - 39 k - 42 = 0 . . . . . (3)$
Solving $(2)$ and $(3)$ , we get $h = 3, k = - 1$
Hence the required pole is $(3, - 1)$ .

Suggest Corrections

Similar questions

9 x + y - 28 = 0

{2 x}^{2} + {2 y}^{2} - 3 x + 5 y - 7 = 0

9 x + y - 28 = 0

2 x^{2} + 2 y^{2} - 3 x + 5 y - 7 = 0

x^{2} + y^{2} + 4 x - 7 = 0

2 x^{2} + 2 y^{2} + 3 x + 5 y - 9 = 0

x^{2} + y^{2} + y = 0

3 x + 5 y + 17 = 0

x^{^{2}} + y^{2} + 4 x + 6 y + 9 = 0

2 x^{2} + 2 y^{2} = 3 x - 5 y + 7

Join BYJU'S Learning Program