Question

If the endpoints of the diameter is $P(3,1)$ and $Q(h,k)$ of the circle $2x^2+2y^2-2x-7y-7=0$, then which of the following is/are correct?

• A. $Q=(-1,2)$
• B. $Q=(-2,\frac{5}{2})$
• C. radius of circle is $\sqrt{\frac{67}{4}}$ units
• D. radius of circle is $\frac{\sqrt{109}}{4}$ units

Answer 1

Answer: (B), (D)

radius of circle is $\frac{\sqrt{109}}{4}$ units

Given: $2x^2+2y^2-2x-7y-7=0$
$\Rightarrow x^2+y^2-x-\frac{7y}{2}-\frac{7}{2}=0g=-\frac{1}{2},f=-\frac{7}{4},c=-\frac{7}{2}$
$C=(-g,-f)=(\frac{1}{2},\frac{7}{4})r=\sqrt{{(-\frac{1}{2})}^2+{(-\frac{7}{4})}^2+\frac{7}{2}}\Rightarrow r=\frac{\sqrt{109}}{4}$

Endpoint of the diameter are $P(3,1)$ and $Q(h,k)$
Centre is the mid point of the two end points of diameter, so
$\frac{h+3}{2}=\frac{1}{2},\frac{k+1}{2}=\frac{7}{4}\Rightarrow h=-2,k=\frac{5}{2}\therefore Q=(-2,\frac{5}{2})$