Question

Centre of circle passing through the points $(4,5),(3,4)$ and $(5,2)$ is

• A. $(\frac{9}{2},\frac{7}{2})$
• B. $(\frac{7}{2},\frac{9}{2})$
• C. $(\frac{7}{2},\frac{7}{2})$
• D. $(\frac{9}{2},\frac{9}{2})$

Answer 1

The correct option is A $(\frac{9}{2},\frac{7}{2})$

Solution:- (A) $(\frac{9}{2},\frac{7}{2})$

Let ${(x-h)}^2+{(y-k)}^2=r^2$ where $(h,k)$ be the centre of circle and $r$ be the radius of circle.

Given that the circle passes through the points $(4,5),(3,4)$ and $(5,2)$, i.e., all the given points will satisfy the equation of circle.

Therefore,

${(4-h)}^2+{(5-k)}^2=r^2$

$16+h^2-8h+25+k^2-10k=r^2$

$h^2+k^2-8h-10k+41=r^2.....\left(1\right)$

${(3-h)}^2+{(4-k)}^2=r^2$

$9+h^2-6h+16+k^2-8k=r^2$

$h^2+k^2-6h-8k+25=r^2.....\left(2\right)$

${(5-h)}^2+{(2-k)}^2=r^2$

$25+h^2-10h+k^2+4-4k=r^2$

$h^2+k^2-10h-4k+29=r^2.....\left(3\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we have

$h^2+k^2-8h-10k+41=h^2+k^2-6h-8k+25$

$2h+2k-16=0$

$\Rightarrow h+k-8=0.....\left(4\right)$

Again from ${eq}^n\left(2\right)\&\left(3\right)$, we have

$h^2+k^2-6h-8k+25=h^2+k^2-10h-4k+29$

$4h-4k-4=0$

$h-k-1=0.....\left(5\right)$

Adding ${eq}^n\left(4\right)\&\left(5\right)$, we have

$(h+k-8)+(h-k-1)=0$

$\Rightarrow 2h-9=0\Rightarrow h=\frac{9}{2}$

Substituting the value of $h$ in ${eq}^n\left(5\right)$, we have

$\frac{9}{2}-k-1=0\Rightarrow k=\frac{7}{2}$

Substituting the value of $h$ and $k$ in ${eq}^n\left(2\right)$, we have

$r^2={\left(\frac{9}{2}\right)}^2+\left(\frac{7}{2}\right)-6\left(\frac{9}{2}\right)-8\left(\frac{7}{2}\right)+25$

$\Rightarrow r^2=\frac{81}{4}+\frac{49}{4}-27-28+25$

$\Rightarrow r^2=\frac{130}{4}-30$

$\Rightarrow r^2=\frac{10}{4}$

$\Rightarrow r=\frac{\sqrt{5}}{2}$

Hence the equation of circle will be-

${(x-\frac{9}{2})}^2+{(y-\frac{7}{2})}^2={\left(\frac{\sqrt{5}}{2}\right)}^2$

Hence the centre of the circle will be $(\frac{9}{2},\frac{7}{2})$ and radius of the circle will be $\frac{\sqrt{5}}{2}$.