The equation of the common tangent of the two touching circles, $y^{2} + x^{2} - 6 x - 12 y + 37 = 0$ and $x^{2} + y^{2} - 6 y + 7 = 0$ is

x + y - 5 = 0

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

x - y + 5 = 0

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

x - y - 5 = 0

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

x + y + 5 = 0

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

Open in App

Solution

The correct option is C $x - y - 5 = 0$
Let $S_{1} \equiv x^{2} + y^{2} - 6 x - 12 y + 37 = 0$ and $S_{2} \equiv x^{2} + y^{2} - 6 y + 7 = 0$
The equation of common tangent of the two circles is $S_{1} - S_{2} = 0$
$\Rightarrow x^{2} + y^{2} - 6 x - 12 y + 37 - (x^{2} + y^{2} - 6 y + 7) = 0$
$\Rightarrow - 6 x + 6 y + 30 = 0$
$\Rightarrow x - y - 5 = 0$

Suggest Corrections

Similar questions

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^{2} - 6 x = 0

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

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

x + y - 8 = 0

Join BYJU'S Learning Program