The equation of the circle having centre at the origin and passing through the point $(3, 4),$ is

x^{2} + y^{2} = 9

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

x^{2} + y^{2} = 16

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

x^{2} + y^{2} = 25

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

x^{2} + y^{2} + 6 x + 8 y + 25 = 0

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

Open in App

Solution

The correct option is C $x^{2} + y^{2} = 25$
Equation of circle with origin as centre :
$x^{2} + y^{2} = r^{2}$
Circle also passes through point $(3, 4)$
$∴ 3^{2} + 4^{2} = 25 = r^{2}$
Hence, equation of circle is $x^{2} + y^{2} = 25.$

Suggest Corrections

Similar questions

Equation of the circle passing through $(2, 3)$ and centre at the origin is _______ .

\sqrt{5}

- - \to O X

Write the equation of the circle passing through (3,4) and touching y-axis at the origin.

(4, 5), (3, 4)

(5, 2)

Join BYJU'S Learning Program