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Q.
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.
Conic Sections
Solution:
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
$4 + 4 + 4 g -4 f+ c = 0$
$\Rightarrow 4 g -4 f+ c + 8 = 0$ $\,...(ii)$
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
$-2 g - 2 f- 7 = 0$
$\Rightarrow 2g + 2f+ 7 = 0$ $\,...(iv)$
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}$
$=\frac{\sqrt{37}}{2}$.