JEE Main 2019. Previous Year Question.
Problem
The equation of a common tangent to the circle $x^2+y^2-6x=0$ and the parabola $y^2=4x$ is:
Key insight. Rather than guessing a tangent to the circle, it’s easier to write the tangent to the parabola in slope form (where the parameters are directly determined by the slope), then impose the circle’s tangency condition on that same line — turning two separate tangency requirements into one equation in a single unknown, the slope $m$.
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Approach
Rewrite the circle’s equation in centre-radius form by completing the square. Write the general tangent to the parabola $y^2=4x$ in slope form, $y=mx+\dfrac1m$. Impose the condition that this line is also tangent to the circle — the perpendicular distance from the circle’s centre to the line equals its radius — and solve for $m$.
Solution
Step 1 — Rewrite the circle in centre-radius form
$$x^2+y^2-6x=0 \implies (x-3)^2+y^2=9$$
Centre $(3,0)$, radius $3$.
Step 2 — Write the tangent to the parabola in slope form
For $y^2=4ax$ with $a=1$ (since $y^2=4x$), a tangent with slope $m$ has the form:
$$y=mx+\frac1m$$
Step 3 — Apply the circle’s tangency condition
Rewrite the tangent line as $mx-y+\dfrac1m=0$. The perpendicular distance from the centre $(3,0)$ to this line must equal the radius $3$:
$$\frac{\left|m(3)-0+\dfrac1m\right|}{\sqrt{m^2+1}} = 3$$
Step 4 — Square both sides
$$\left(3m+\frac1m\right)^2 = 9(m^2+1)$$
$$9m^2+6+\frac{1}{m^2} = 9m^2+9$$
Step 5 — Simplify
The $9m^2$ terms cancel:
$$6+\frac{1}{m^2} = 9 \implies \frac{1}{m^2}=3 \implies m^2=\frac13 \implies m=\pm\frac{1}{\sqrt3}$$
Step 6 — Write the tangent line using m = 1/√3
$$y = \frac{x}{\sqrt3}+\sqrt3$$
Multiplying through by $\sqrt3$:
$$\sqrt3y = x+3$$
Answer
$$\sqrt3y = x+3$$
Common mistakes
- Starting with the tangent to the circle instead of the parabola. A tangent to a circle in slope form carries an extra parameter (the point of contact, or equivalently a $\pm$ under a square root) that a parabola’s tangent in slope form doesn’t — starting from the parabola’s simpler one-parameter form (just $m$) keeps the algebra far more manageable.
- Forgetting the absolute value when setting up the distance formula, then losing the $\pm$ options for $m$. Since $m^2=\dfrac13$ has two real solutions, both $m=\dfrac{1}{\sqrt3}$ and $m=-\dfrac{1}{\sqrt3}$ give valid common tangents — matching against the given options (rather than assuming only one sign is “the” answer) is what pins down which one is listed.
Practise next
- Find the equation of a common tangent to the circle $x^2+y^2-4x=0$ and the parabola $y^2=8x$, using the same slope-form-and-distance approach.
Show answer
$x=0$ — and the slope form cannot find it. A tangent to $y^2=8x$ with slope $m$ is $y=mx+\tfrac2m$. For it to touch the circle (centre $(2,0)$, radius $2$) needs $\dfrac{|2m+2/m|}{\sqrt{1+m^2}}=2$.
Squaring and simplifying leaves $m^2+1=m^2$, which has no solution at all. So no common tangent has a finite slope.
The vertical line has to be checked separately: $x=0$ is the tangent to the parabola at its vertex, and it is at distance $2$ from $(2,0)$, so it touches the circle too. In fact the two curves touch each other at the origin, and $x=0$ is the only common tangent. The slope form silently excludes vertical lines — worth remembering whenever it returns nothing.
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