Question

The equation of the circle which passes through the points $(2,3)$ and $\left(4.5\right)$ and the centre lies on the straight line $y-4x+3=0$, is

• A. $x^2+y^2+4x-10y+25=0$
• B. $x^2-y^2-4x-10y+25=0$
• C. $x^2+y^2-4x-10y+16=0$
• D. $x^2+y^2-14y+8=0$

Answer 1

Answer: (B)

$x^2-y^2-4x-10y+25=0$

Centre lies on the line $y-4x+3=0$

Let $x=h$

$\Rightarrow y=4h-3$

So the center is of the form $(h,4h-3)$

Distance of centre from $(2,3)$ and $(4,5)$ will be equal

$\Rightarrow {(h-2)}^2+{(4h-3-3)}^2={(h-4)}^2+{(4h-3-5)}^2\Rightarrow h=2$

So the centre is $(2,5)$

$r=\sqrt{{(2-2)}^2+{(5-3)}^2}=2$

So the equation of the circle is

${(x-2)}^2+{(y-5)}^2=2^2{x}^2+y^2-4x-10y+25=0$