Question

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

Answer 1

Given

$x^2+y^2-4x-6y+11=0$

$\Rightarrow (x-2{)}^2+(y-3{)}^2=2$

$\to (2,3)$ is center of circle and P(2,3)and Q(3,4) lies on same line

$E{q}^n$ of line passing through P & Q is y = x+1.

Here $m=1=tan\theta \Rightarrow \theta =\pi /4$

$R_1[\sqrt{2}cos\theta +2,3+\sqrt{2}sin\theta ]=3,4$

$R_2[\sqrt{2}cos\theta +2,3-\sqrt{2}sin\theta ]=(1,2)$

$\therefore$ co-ordinate of other end of diameter is (1,2)