If one end of a diamter of the circle $3 x^{2} + 3 y^{2} - 9 x + 6 y + 5 = 0$ is (1, 2) the other end is

(2, 1)

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

(2, 4)

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

(2, - 4)

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

(- 4, 2)

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

Open in App

Solution

The correct option is C $(2, - 4)$
Let $A (1, 2)$ be one end of diameter and B(x,y) be the other end

Center of circle = $(\frac{9}{2 \times 3}, \frac{- 6}{2 \times 3}) i . e . (\frac{- g}{a}, \frac{- f}{a})$

$= (\frac{3}{2}, - 1)$

WKT C is the midpoint of AB

$⟹ (\frac{3}{2}, - 1) = (\frac{x + 1}{2}, \frac{y + 2}{2})$

$⟹ x = 2, y = - 4$

$B = (2, - 4)$

Suggest Corrections

Similar questions

3 x^{2} + 3 y^{2} - 9 x + 6 y + 5 = 0

(1, 2)

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

(3, 4)

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

x^{2} + y^{2} + 4 x - 8 y + 5 = 0

(2, 1)

3 x^{2} + 3 y^{2} + λ x y + 9 x + (λ - 6) y + 3 = 0

Join BYJU'S Learning Program