Question

If the co-ordinates at one end of a diameter of the circle $x^2+y^2-8x-4y+c=0$ are $(11,2)$ the coordinates at the other end are

• A. $(3,2)$
• B. $(-3,-2)$
• C. $(-3,2)$
• D. $(3,-2)$

Answer 1

The correct option is C $(-3,2)$

$x^2+y^2-8x-4y+c=0$

$(11,2)\to$ first coordinates centre $(4,2)$

Coordinates which lies on the circle is $(11,2)$

Let the coordinates on the other end of the diameter be $(x_1,y_1)$

then coordinates of mid point of these $2$ points is $(4,2)$

$\Rightarrow \frac{x_2+11}{2}=4$

$x_1=8-11=-3$

$x_1=-3$

$\frac{y_1+2}{2}=2$

$y_1=4-2=2$

$y_1=2$

Hence the coordinates of the point on the other end of the diameter $(-3,2)$