Find the equation of the circle which passes through the point $(1, 1)$ & which touches the $x^{2} + y^{2} + 4 x - 6 y - 3 = 0$ at the point $(2, 3)$ on it.

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

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

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

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

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

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

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

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

Open in App

Solution

The correct option is A $x^{2} + y^{2} + x - 6 y + 3 = 0$

Correction(in 3rd last step): $x^{2} + y^{2} ‘ ‘ + " x - 6 y = \frac{9}{4} - \frac{1}{4} + 4 - 9$
Change $- x$ to $+ x$ in following steps

Suggest Corrections

Similar questions

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} - 2 x - 6 y + 6 = 0

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

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

x^{2} + y^{2} - 8 x + 6 y - 5 = 0

(- 2, - 7)

x^{2} + y^{2} + 8 x + 10 y - 7 = 0

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

Join BYJU'S Learning Program