The other end of the diameter through the point (-1,1) on the circle x2+y2-6 x+4 y-12=0 is

Question

The other end of the diameter through the point (-1,1) on the circle x2+y2-6 x+4 y-12=0 is (A) (-7,5) (B) (-7,-5) (C) (7,-5) (D) (7, 5)

Tardigrade Admin · Accepted Answer

Given, circle is x2+y2-6 x+4 y-12=0 Centre of this circle is (3,-2) Let other end of the diameter is (α, β). ∴ (α-1/2)=3, (β+1/2)=-2 ⇒ α=7, β=-5 ∴ Other end of the diameter is (7,-5).

Q. The other end of the diameter through the point $(- 1, 1)$ on the circle $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ is

(-7,5)

(-7,-5)

(7,-5)

(7, 5)

Given, circle is $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$
Centre of this circle is $(3, - 2)$
Let other end of the diameter is $(α, β)$ .
$∴ \frac{α - 1}{2} = 3, \frac{β + 1}{2} = - 2$
$\Rightarrow α = 7, β = - 5$
$∴$ Other end of the diameter is $(7, - 5)$ .

Q. The other end of the diameter through the point (−1,1) on the circle x2+y2−6x+4y−12=0 is