Aspire Faculty ID #16555 · Topic: NIMCET 2008 · Just now
NIMCET 2008

Equation of the common tangent touching the circle $(x-3)^2+y^2=9$ and the parabola $y^2=4x$ above the $x$-axis is

Solution

Let tangent be $y=mx+c$, $m>0$. Tangency with $y^2=4x$ gives $c=\dfrac1m$. Tangency with the circle gives $|3m-c|=3\sqrt{m^2+1}$. Solving gives $m=\dfrac1{\sqrt3},\ c=\sqrt3$. Equation: $\sqrt3y=x+3$ Answer: $\boxed{\sqrt3y=x+3}$

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