Question

The equation of the circle which passes through $(8,16)$ and touches the line $4x-3y=64$ at $(16,0)$ is

• A. $x^2+y^2-16x-12y=0$
• B. $x^2+y^2-12x-16y=0$
• C. $x^2+y^2-16x-12y=100$
• D. $x^2+y^2-12x-16y=100$

Answer 1

The correct option is A $x^2+y^2-16x-12y=0$

Let the equation of the circle is
$(x-h{)}^2+(y-k{)}^2=r^2$
$\because 4x-3y=64$ is the tangent to the circle at $(16,0)$
$\therefore$ Perpendicular line to $4x-3y=64$ at $(16,0)$ is $3x+4y=48$

Center $(h,k)$ lies on the line $3x+4y=48$
$\Rightarrow h=16-\frac{4}{3}k\cdots \left(1\right)$
$\text{and}\frac{|4h-3k-64|}{5}=r\Rightarrow r=\frac{|5k|}{3}\cdots \left(2\right)$

$\because$ Circle passes through $(8,16)$
$\therefore (8-h{)}^2+(16-k{)}^2=r^2$
From $\left(1\right)$ and $\left(2\right),$ the equation becomes
${(\frac{4}{3}k-8)}^2+(16-k{)}^2=\frac{25}{9}{k}^2\therefore k=6$
$\Rightarrow r=10,h=8$