The equation of the pair of tangents at $(0, 1)$ to the circle $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is

3 (x^{2} - y^{2}) + 4 x y - 4 x - 6 y + 3 = 0

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

3 y^{2} + 4 x y - 4 x - 6 y + 3 = 0

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

3 x^{2} + 4 x y - 4 x - 6 y + 3 = 0

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

3 (x^{2} + y^{2}) + 4 x y - 4 x - 6 y + 3 = 0

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

Open in App

Solution

The correct option is D $3 y^{2} + 4 x y - 4 x - 6 y + 3 = 0$
Let
$S \equiv x^{2} + y^{2} - 2 x - 6 y + 6 = 0$
and
$P (x_{1}, y_{1}) = (0, 1)$
$S_{1} = x_{1}^{2} + y_{1}^{2} - 2 x_{1} - 6 y_{1} + 6 = 0$
$= (0)^{2} + (1)^{2} - 2 (0) - 6 (1) + 6$
$= 1 - 6 + 6 = 1$

$T = x \cdot x_{1} + y \cdot y_{1} - (x + x_{1}) - 3 (y + y_{1}) + 6$
$= x \cdot (0) + y (1) - (x + 0) - 3 (y + 1) + 6$
$= 0 + y - x - 3 y - 3 + 6$
$= - x - 2 y + 3$

$\Rightarrow T^{2} = (- x - 2 y + 3)^{2}$
$= (- x - 2 y)^{2} + 9 + 6 (- x - 2 y)$
$= x^{2} + 4 y^{2} + 4 x y + 9 - 6 x - 12 y$

$∴$ Equation of the pair of tangents

$S \cdot S_{1} = T^{2}$
$(x^{2} + y^{2} - 2 x - 6 y + 6) (1)$
$= x^{2} + 4 y^{2} + 4 x y - 6 x - 12 y + 9$
$\Rightarrow 3 y^{2} + 4 x y - 4 x - 6 y + 3 = 0$

Suggest Corrections

Similar questions

Find the equations of transverse common tangents for two circles

$x^{2} + y^{2} + 6 x - 2 y + 1 = 0, x^{2} + y^{2} - 2 x - 6 y + 9 = 0$

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

2^{nd}

4^{th}

The diameters of a circle are along $2 x + y - 7 = 0$ and $x + 3 y - 11 = 0$ Then the equation of this circle, which also passes through (5,7) is

x^{2} + y^{2} + 4 x - 6 y - 3 = 0

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

x^{2} + y^{2} - 4 x - 6 y - 8 = 0

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

Join BYJU'S Learning Program