Question

Given a curve $C.$ Suppose that the tangent line at $P(x,y)$ on $C$ is perpendicular to the line joining $P$ and $Q(1,0).$ If the line $2x+3y-15=0$ is tangent to the curve $C,$ then the curve $C$ denotes

• A. a circle touching the $x-$axis
• B. a circle touching the $y-$axis
• C. a circle with $y-$intercept equal to $4\sqrt{3}$ units
• D. a parabola with axis of symmetry parallel to $y-$axis

Answer 1

The correct option is C a circle with $y-$intercept equal to $4\sqrt{3}$ units


Let slope of the curve $C$ at $P(x,y)$ be $m.$
Then, $\left(\frac{y-0}{x-1}\right)m=-1$
$\Rightarrow y\frac{dy}{dx}=1-x$
$\Rightarrow \frac{y^2}{2}=x-\frac{x^2}{2}+C^{\prime }$
$\Rightarrow x^2+y^2-2x=C$ $(C=2C^{\prime })$
This is equation of a circle with centre $(1,0)$
Since $2x+3y-15=0$ is tangent to the circle,
$\therefore$ Perpendicular distance from $(1,0)$ on the line $=$ radius
$\Rightarrow \left|\frac{2-15}{\sqrt{13}}\right|=\sqrt{1+C}$
$\Rightarrow C+1=13$
$\Rightarrow C=12$
Hence, equation of the circle is $x^2+y^2-2x=12.$
It represents the circle with $y-$intercept $=2\sqrt{f^2-c}=4\sqrt{3}$