The equation of circle passing through (4, 5) and having the centre at (2, 2), is

x^{2} + y^{2} + 4 x + 4 y - 5 = 0

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

x^{2} + y^{2} - 4 x - 4 y - 5 = 0

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

x^{2} + y^{2} - 4 x = 13

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

x^{2} + y^{2} - 4 x - 4 y + 5 = 0

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

Open in App

Solution

The correct option is B $x^{2} + y^{2} - 4 x - 4 y - 5 = 0$
Centre (2, 2) and
R = $\sqrt{(4 - 2)^{2} + (5 - 2)^{2}} = \sqrt{13}$
Hence required equation is $x^{2} + y^{2} - 4 x - 4 y - 5 = 0.$

Suggest Corrections

Similar questions

(5, 2)

(1, - 1)

(2, 2)

(4, 5)

\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1

(0, 3)

\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1,

(0, 3)

Join BYJU'S Learning Program