The centres of the circles $x^{2} + y^{2} = 1,$ $x^{2} + y^{2} + 6 x - 2 y = 1$ and $x^{2} + y^{2} - 12 x + 4 y = 1$ are

same

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

Collinear

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

Non–collinear

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 Collinear
Centre are (0, 0), (-3, 1) and (6, -2) and a line passing through any two points say (0, 0) and (-3, 1) is $y = - \frac{1}{3} x$ and point (6, -2) lies on it. Hence points are collinear.

Suggest Corrections

Similar questions

x^{2} + y^{2} = 1, x^{2} + y^{2} + 6 x - 2 y = 1

x^{2} + y^{2} - 12 x + 4 y = 1

x^{2} + y^{2} - 4 x - 6 y + 5 = 0

x^{2} + y^{2} - 2 x - 4 y - 1 = 0

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

x^{2} + y^{2} - 4 x - 6 y + 5 = 0, x^{2} + y^{2} - 2 x - 4 y - 1 = 0, x^{2} + y^{2} - 6 x - 2 y = 0

x^{2} + y^{2} - 8 x - 12 y = 1, x^{2} + y^{2} - 6 x - 4 y = 1, x^{2} + y^{2} - 5 x - 7 y = 1

Join BYJU'S Learning Program