Question

The equation of the circle passing through the points $(4,1),(6,5)$ and having the centre on the line $4x+y-16=0$ is

• A. $x^2+y^2-6x-8y+15=0$
• B. $15(x^2+y^2)-94x+18y+55=0$
• C. $x^2+y^2-4x-3y=0$
• D. $x^2+y^2+6x-4y=0$

Answer 1

Answer: (A)

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

Given points,

$(4,1),(6,5)$

equation of circle $(x-h{)}^2+(y-k{)}^2=r^2$

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

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

solving the above 2 equations, we get,

$h+2k=11$....(3)

given, $4h+k=16$.....(4)

solving the above 2 equations, we get,

$h=3,k=4$

substituting the above values in (1), we get,

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

$\therefore r=\sqrt{10}$

Hence, the equation is,

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

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