Question

A circle touching the $x$-axis at $(3,0)$ and making an intercept of length $8$ on the $y$-axis passes through the point :

• A. $(3,10)$
• B. $(3,5)$
• C. $(1,5)$
• D. $(2,3)$

Answer 1

The correct option is A $(3,10)$

Let $r$ be the radius of the circle.
Equation of required circle is
$(x–3{)}^2+(y\pm r{)}^2=r^2$
$\Rightarrow x^2–6x+9+y^2\pm 2ry+r^2=r^2$
$\Rightarrow x^2+y^2–6x\pm 2ry+9=0\dots \left(1\right)$

Length of $y$-intercept of eqn. $\left(1\right)$ is given by
$2\sqrt{f^2-c},f=\pm r$
$\Rightarrow 8=2\sqrt{r^2-9}\Rightarrow 16=r^2–9\Rightarrow r=5$

So, required circle is
$x^2+y^2–6x+10y+9=0\dots \left(2\right)x^2+y^2–6x–10y+9=0...\left(3\right)$
By putting all the points in these two equations, we get
$(3,10)$ satisfies eqn. $\left(3\right)$.