Find the equation of the circle that passes through the points $(1.0), (- 1, 0)$ and $(0, 1)$

x^{2} + y^{2} = 1

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

x^{2} - y^{2} = 1

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

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

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

None of these

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

Open in App

Solution

The correct option is B $x^{2} + y^{2} = 1$
$∠ A O B = 90^{o}$
$∴$ Circle will be formed with AB as diameter
i.e.,
$(x + 1) (x - 1) + (y - 0) (y - 0) = 0$
$x^{2} - 1 + y^{2} = 0$
$x^{2} + y^{2} = 1$ .

Suggest Corrections

Similar questions

\frac{x^{2}}{4} + y^{2} = 1

\frac{x^{2}}{a^{2}} + y^{2} = 1

^{'} a^{'}

(1, 1)

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

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

(1 + y^{2}) d x - x y d y = 0

The equation of the circle concentric with $x^{2} - 3 x + 4 y - c = 0$ and passing through (-1, -2) is

Join BYJU'S Learning Program