Question

A circle passes through the points $(2,-2)$, $(3,4)$ and has its centre on the line $2x+2y=7$. Find its centre and radius.

• A. $\left(1,\frac{5}{2}\right),\frac{\sqrt{37}}{2}$
• B. $\left(5,1\right),\sqrt{37}$
• C. $\left(1,5\right),\sqrt{37}$
• D. $\left(\frac{5}{2},1\right),\frac{\sqrt{37}}{2}$

Answer 1

Answer: (D)

Let the equation of the circle be
$x^2+y^2+2gx+2fy+c=0$ $...(i)$
As $(2,-2)$, $(3,4)$ lie on the circle, we have
$4+4+4g-4f+c=0$
$\Rightarrow 4g-4f+c+8=0$ $...(ii)$
and $9+16+6g+8f+c=0$
$\Rightarrow 6g+8f+c+25=0$ $...(iii)$
Since the centre $(-g,-f)$ lies on
$2x+2y-7=0$, therefore, we get
$-2g-2f-7=0$
$\Rightarrow 2g+2f+7=0$ $...(iv)$
Subtracting $(ii)$ from $(iii)$, we get
$2g+12f+17=0$ $...(v)$
Subtracting $(iv)$ from $(v)$, we get $10f+10=0$
$\Rightarrow f=-1$
Putting value of $f$ in $(iv)$, we get $2g+5=0$
$\Rightarrow g=-\frac{5}{2}$,
and then $(ii)$ gives $c=-2$.
Also, we note that $g^2+f^2-c=\frac{25}{4}+1+2=\frac{37}{4}$, which is positive.
Its centre is $\left(\frac{5}{2},1\right)$
and radius = $\sqrt{\frac{25}{4}+1+2}$
$=\frac{\sqrt{37}}{2}$.