Question

The equation of the circle passing through $(4,1)$ and $(6,5)$ and having center of the line

• A. $x^2+y^2-6x-8y+15=0$
• B. $x^2+y^2-6x-8y-15=0$
• C. $x^2+y^2-6x-8y-11=0$
• D. $x^2+y^2-6x-8y+11=0$

Answer 1

The correct option is A $x^2+y^2-6x-8y+15=0$

Let the equation of the required circle be $(x-h{)}^2+(y-k{)}^2=r^2$ …………$\left(1\right)$

It is given that the circle passes through $(4,1)$ and $(6,5)$

$\therefore (4-h{)}^2+(1-k{)}^2=r^2$ and $(6-h{)}^2+(5-k{)}^2=r^2$

$\Rightarrow (4-h{)}^2+(1-k{)}^2=(6-h{)}^2+(5-k{)}^2$

on expanding we get,

$16-8h+h^2+1-2k+k^2=36-12h+h^2+25-10k+k^2$

$16-8h+1-2k=36-12h+25-10k$

$4h+8k=44$.

Dividing throughout by $4$, we get,

$h+2k=1$ ……….$\left(2\right)$

It is given that the centre of the circle lies on the line $4x+y=16$

$\therefore 4h+k=16$ ………….$\left(3\right)$

Solving $\left(2\right)$ & $\left(3\right)$

$4h+k=16$

$4h+8k=44$

_____________

$-7k=-25$

$k=4$

_____________

Substitute for k in equation $\left(1\right)$

$h+2\left(4\right)=11$

$h=3$

$\therefore h=3$ & $k=4$ putting in equation $\left(1\right)$

$(4-3{)}^2+(1-4{)}^2=r^2$

$r=\sqrt{10}$

Hence, $(x-3{)}^2+(y-4{)}^2=(\sqrt{10}{)}^2$

$\therefore x^2+y^2-6x-8y+15=0$.