Question

One end of a diameter of the circle $x^2+y^2-6x+5y-7=0$ is (-1, 3). Find the co-ordinates of the other end.

Answer 1

We have,

Equation of given circle is

$x^2+y^2-6x+5y-7=0$

Then, centre of circle

Comparing that,

$x^2+y^2+2gx+2fy+c=0$

$(-g,-f)=(3,\frac{-5}{2})$

Then, centre of $(x,y)=(3,\frac{-5}{2})$

One side of point $(x_1,y_1)=(-1,3)$

Let the end point of $(x_2,y_2)$

$(x,y)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$(3,\frac{-5}{2})=(\frac{-1+x_2}{2},\frac{3+y_2}{2})$

Then,

$\frac{-1+x_2}{2}=3$

$x_2-1=6$

$x_2=5$

$\frac{-5}{2}=\frac{3+y_2}{2}$

$y_2=3+5=8$

$y_2=8$

Then, end point of diameter$(x_2,y_2)=(5,8)$

Hence, this is the answer.