Question

Show that the points $(5,5)$, $(6,4)$, $(-2,4)$ and $(7,1)$ Concyclic find its equation, centre and radius.

Answer 1

consider the problem

If four points are concyclic then $4^{th}$ point must satisfy equation of circle passing through other three points.

Let, $x^2+y^2+2gx+2fy+c=0$ be equation of circle

$(5,6),(6,4),and(7,1)$ must satisfy circle

$\begin{array}{c}\left(5,5\right)\to 25+25+10g+10f+c=0\\ \Rightarrow 10g+10f+c=-50\left(i\right)\end{array}$

$\begin{array}{c}\left(6,4\right)\to 36+16+12g+8f+c=0\\ \Rightarrow 12g+8f+c=-52\left(ii\right)\end{array}$

$\begin{array}{c}\left(7,1\right)\to 49+1+14g+2f+c=0\\ \Rightarrow 14g+2f+c=-50\left(iii\right)\end{array}$

And $\left(ii\right)-\left(i\right)$

$\Rightarrow 2g-2f=-2\dots \left(iv\right)$

$\left(iii\right)-\left(ii\right)$

$\Rightarrow 2g-6f=2\dots \left(v\right)$

$\left(iv\right)-\left(v\right)$

$\begin{array}{c}2g-2f-2g+6f=-2-2\\ \Rightarrow 4f=-4\\ \Rightarrow f=-1\end{array}$

Put in $\left(iv\right)$

$\begin{array}{c}2g+2=-2\\ \Rightarrow g=-2\end{array}$

$10g+10f+c=-50$

Put in $\left(i\right)$

$\begin{array}{c}-20-10+c=-50\\ \Rightarrow c=-20\end{array}$

Therefore, equation is $x^2+y^2-4x-2y-20=0$

Now, put $(-2,4)$ in $R.H.S$

$4+16+8-8-20=28-28=0$

Therefore, $(-2,4)$ satisfies circle passing through $(5,5),(6,4)and(7,1)$

Hence, $(5,5),(6,4),(-2,4)and(7,1)$ are concyclic

Center is $(2,1)$

And Radius is

$\begin{array}{c}=\sqrt{4+1-\left(-20\right)}\\ =\sqrt{25}\\ =5\end{array}$