Question

Find the equation of the circle passing through the points $\left(2,3\right)$ and $(-1,1)$ and whose centre is on the line $x-3y-11=0$ .

Answer 1

The equation of circle
$(x-h)62+(y-k{)}^2=r^2...(i)$
According to question, circle passes through the point $(2,3)$ and $(-1,1)$
Thus $(2-h)62+(3-k{)}^2=r^2$
$\Rightarrow 4+h^2-4h+9+k^2-6k=r^2$
and $(-1-h{)}^2+(1-k{)}^2=r^2$
$\Rightarrow 1+h^2+2h+1+k^2-2k=r^2$
Again $h^2+k^2-4h-6k+12=r^2..\left(ii\right)$
and $h^2+k^2+2h-2k+2-r^2..\left(iii\right)$
Since centre of circle $(h,k)$ lie on line
$x-3y-11=0$
Thus $h-3k-11-0$
$\Rightarrow h-3k-\mathrm{11...}\left(iv\right)$
Subtraction eq $\left(iii\right)$ from $\left(ii\right)$,
$\frac{\begin{array}{c}{h}^2+k^2-4h-6k+12=r^2\\ h^2+k^2+2h-2k+2=r^2\end{array}\begin{array}{c}---+--\end{array}}{\begin{array}{c}-6h-4k+11=0\\ -6h-4k=-11\end{array}\begin{array}{c}6h+4k=11\end{array}...\left(iv\right)}$
Multiply eq. $\left(iv\right)$ by $4$ and eq. $\left(v\right)$ by $3$ and then adding
$4h-12k=44$
$18h+12k=33$
$22h=77$
$\Rightarrow h=\frac{77}{22}$
$\therefore h=\frac{7}{2}$
Put the value of $h$ in equation $\left(iv\right)$,
$\frac{7}{2}-3k=11$
$\Rightarrow 3k=\frac{7}{2}-11$
$\Rightarrow 3k=\frac{7-22}{2}$
$\Rightarrow 3k=-\frac{15}{2}$
$\Rightarrow k=-\frac{15}{2\times 3}$
$\therefore K=-\frac{5}{2}$
Put the values of $h$ and $k$ in equation $\left(ii\right)$,
${\left(\frac{7}{2}\right)}^2+{(-\frac{5}{2})}^2-4\times \frac{7}{2}-6\times (-\frac{5}{2})+12=r^2$
$\Rightarrow \frac{49}{4}+\frac{25}{4}-14+15+13=r^2$
$\Rightarrow \frac{74}{4}+14=r^2$
$\Rightarrow \frac{74+56}{4}=r^2$
$\Rightarrow r62=\frac{130}{4}$
$\Rightarrow r=\sqrt{\frac{130}{4}}$
Put teh values of $h,k$ and $r$ in equation $\left(i\right)$
${(x-\frac{7}{2})}^2+{[y-(-\frac{5}{2})]}^2={\left(\sqrt{\frac{130}{4}}\right)}^2$
$\Rightarrow {(x-\frac{7}{2})}^2+{(y+\frac{5}{2})}^2=\frac{130}{4}$
$\Rightarrow x^2+\frac{49}{4}-2\times x\times \frac{7}{2}+y^2+\frac{25}{4}+2\times \frac{5}{2}\times y=\frac{130}{4}$
$\Rightarrow x^2+y^2-7x+5y+\frac{49}{4}+\frac{25}{4}=\frac{130}{4}$
$\Rightarrow x^2+y^2-7x+5y+\frac{74}{4}-\frac{130}{4}=0$
$\Rightarrow x^2+y^2-7x+5y+\frac{-56}{4}=0$
$\Rightarrow x^2+y^2-7x+5y+-14=0$
Thus, required equation of circle
$x^2+y^2-7x+5y+-14=0$