Equation of the normal to the circle $x^{2} + y^{2} - 4 x + 4 y - 17 = 0$ which passes through $(1, 1)$ is

3 x + 2 y - 5 = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3 x + y - 4 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 x + 2 y - 2 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3 x - y - 8 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $3 x + 2 y - 5 = 0$
$x^{2} + y^{2} - 4 x + 4 y - 17 = 0$

Centre $(\frac{+ y}{2}, \frac{- y}{2})$ i.e., $(2, - 2)$

We know normal to a circle pass through the centre

so we have 2 points $(2, - 2)$ and $(1, 1)$

$∴$ Equation $= (y - 1) = (\frac{- 2 - 1}{2 - 1}) (x - 1)$

$\Rightarrow$ $3 x + y - 4 = 0$

Suggest Corrections

Similar questions

x^{2} + y^{2} - 6 x + 2 y - 8 = 0

The common chord of the circle $x^{2} + y^{2} + 4 x + 1 = 0$ and $x^{2} + y^{2} + 6 x + 2 y + 3 = 0$ is

2 x - y - 2 = 0

(3, - 2)

(- 2, 0)

x + y = 5, 2 x + 2 y = 10

x - y = 8, 3 x - 3 y = 16

2 x + y - 6 = 0, 4 x - 2 y - 4 = 0

2 x - 2 y - 2 = 0, 4 x - 4 y - 5 = 0

Join BYJU'S Learning Program