Question

If the coordinates at one end of a diameter of the circle $x^2+y^2-8x-4y+c=0$ are $(-3,2)$, then the coordinates at the other end are

• A. $(5,3)$
• B. $(6,2)$
• C. $(1,-8)$
• D. $(11,2)$

Answer 1

Answer: (D)

The centre of the given circle is $C\equiv (4,2)$
Let $A\equiv (-3,2)$
If $(\alpha ,\beta )$ are the coordinates of the other end of the diameter, then, as the middle ploint of the diameter is the centre,
$\therefore \frac{\alpha -3}{2}=4$ and $\frac{\beta +2}{2}=2\Rightarrow \alpha =11,\beta =2$
Thus, the coordinates of the other end of diameter are $(11,2)$