Question

If (4,1) is one of extremity of a diameter of the circle x^2+y^2-2x+6y-15=0 find the other extremity

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{(4,1) is one of the extremity of a diameter of the circle}$

$\mathsf{x^2+y^2-2x+6y-15=0}$

$\underline{\textsf{To find:}}$

$\textsf{The other extremity}$

$\underline{\textsf{Solution:}}$

$\textsf{Let the other extremity be (a,b)}$

$\mathsf{x^2+y^2-2x+6y-15=0}$

$\textsf{Comparing it with}\;\mathsf{x^2+y^2+2gx+2fy+c=0}$

$\textsf{we get 2g=-2, 2f=6}$

$\implies\textsf{g=-1, f=3}$

$\therefore\textsf{Centre is (-g,-f)=(1,-3)}$

$\textsf{We know that}$

$\textsf{"Centre of circle is the midpoint of extremities of diameter"}$

$\implies\mathsf{(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})=(1.-3)}$

$\implies\mathsf{(\dfrac{4+a}{2},\dfrac{1+b}{2})=(1.-3)}$

$\textsf{Equating corresponding coordinates on bothsides, we get}$

$\begin{array}{c|c}\dfrac{4+a}{2}=1&\dfrac{1+b}{2}=-3\\&\\4+a=2&1+b=-6\\&\\a=-2&b=-7\end{array}}$

$\therefore\textsf{The other extremity is (-2,-7)}$