Question

The line $4x-3y=-12$ is tangent at the point $(-3,0)$ and the line $3x+4y=16$ is tangent at the point $(4,1)$ to a circle. The equation of the circle is

• A. $x^2+y^2-2x+6y-15=0$
• B. $x^2+y^2-2x+6y-20=0$
• C. $x^2+y^2+2x+6y-15=0$
• D. $x^2+y^2-2x-6y-15=0$

Answer 1

Answer: (A)

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

Clearly, centre $C$ of the required circle is the point of intersection of the normals at $A$ and $B$.
The equation of a line through $A(-3,0)$ and the perpendicular to $4x-3y=-12$ is
$y-0=\frac{-3}{4}(x+3)\Rightarrow 3x+4y+9=0...\left(i\right)$
Similarly, the equation of a line through $B(4,1)$ and perpendicular to $3x+4y=16$ is
$y-1=\frac{4}{3}(x-4)\Rightarrow 4x-3y-13=0...\left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we get $x=1$, $y=-3$.
So, the coordinates of $C$ are $(1,-3).$
$\therefore$ Radius = $CA=\sqrt{16+9}=5.$
Hence, the equation of the required circle is
$(x-1{)}^2+(y+3{)}^2=5^2\text{or,}{x}^2+y^2-2x+6y-15=0$.