Eqaution of the tangent to $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ at $(- 1, 1)$

4 x - 3 y + 7 = 0

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

4 x + 3 y - 7 = 0

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

3 x - 4 y + 2 = 0

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

x - 2 y + 4 = 0

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

Open in App

Solution

The correct option is A $4 x - 3 y + 7 = 0$
Equation of tangent to $x^{2} + y^{2} - 6 x + 6 y - 12 = 0$ at (1,1) is $T = 0$
$x x_{1} + y y_{1} - 3 (x + x_{1}) + 2 (y + y_{1}) - 12 = 0$
$- x + y - 3 x + 3 + 2 y + 2 - 12 = 0$
$y - 3 y - 4 x - 7 = 0$
$\Rightarrow 4 x + 7 = 3 y$

Suggest Corrections

Similar questions

x^{2} + y^{2} - 6 x + 4 y = 12

4 x - 3 y + 7 = 0

x^{2} + y^{2} - 6 x + 4 y - 12 = 0

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

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

y - 1 = 0

4 x - 3 y - 9 = 0

x^{2} + y^{2} - 4 x - 10 y + 28 = 0

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

x - 1 = 0, 3 x + 4 y - 21 = 0

x + 2 y - 3 = 0, 3 x + 4 y - 7 = 0, 2 x + 3 y - 4 = 0, 4 x + 5 y - 6 = 0

x^{2} + y^{2} - 6 x + 4 y - 12 = 0

4 x + 3 y + 5 = 0

Join BYJU'S Learning Program