If one end of a diameter of the circle $x^{2} + y^{2} - 4 x - 6 y + 11 = 0$ be (3, 4), then the other end is

(0, 0)

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

(1, 1)

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

(1, 2)

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

(2, 1)

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

Open in App

Solution

The correct option is C (1, 2)
Centre is (2, 3). One end is (3, 4).
$P_{2}$ divides the join of $P_{1}$ and O in ratio of 2 : 1.
Hence $P_{2} i s (\frac{4 - 3}{2 - 1}, \frac{6 - 4}{2 - 1}) \equiv (1, 2) .$

Suggest Corrections

Similar questions

x^{2} + y^{2} - 4 x - 6 y + 11 = 0

x^{2} + y^{2} - 4 x - 6 y + 11 = 0

x^{2} + y^{2} - 4 x - 6 y + 11 = 0

(3, 4)

x^{2} + y^{2} - 4 x - 6 y + 11 = 0

(3, 4)

x^{2} + y^{2} - 4 x - 6 y + 11 = 0

(3, 4)

Join BYJU'S Learning Program