If the Line ax + y = c Touches Both the Curves x² + y² = 1 and y² = 4√2x, Then |c| Is Equal To

JEE Main 2019 — 10 April, Evening Shift. Previous Year Question.

Problem

If the line $ax+y=c$ touches both the curves $x^2+y^2=1$ and $y^2=4\sqrt2x$, then $|c|$ is equal to:

Key insight. Two separate tangency conditions on the same line and unknowns $a,c$ give exactly two equations — one from the circle (perpendicular distance from centre equals radius) and one from the parabola (matching the line’s slope-intercept form against the standard tangent formula). Combining them algebraically eliminates $a$ entirely, leaving a clean equation in $c$ alone.

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Approach

Apply the circle’s tangency condition first: the perpendicular distance from the origin to the line $ax+y-c=0$ equals the circle’s radius, $1$. Then compare the given line, rewritten as $y=-ax+c$, against the standard slope-form tangent to the parabola $y=mx+\dfrac{A}{m}$ (with $A=\sqrt2$ here), which identifies $m=-a$ and $c=\dfrac{\sqrt2}{m}$. Combine both relationships to eliminate $a$ and solve for $c$.

Solution

Step 1 — Apply the circle’s tangency condition

The line $ax+y-c=0$ is tangent to $x^2+y^2=1$ (centre origin, radius $1$), so the perpendicular distance from the origin equals $1$:

$$\frac{|{-c}|}{\sqrt{a^2+1}} = 1 \implies c^2 = a^2+1 \quad \text{…(i)}$$

Step 2 — Identify the parabola’s parameter

$y^2=4\sqrt2x$ matches $y^2=4Ax$ with $A=\sqrt2$.

Step 3 — Match the line against the parabola’s tangent-in-slope-form

The standard tangent to $y^2=4Ax$ in slope form is $y=mx+\dfrac{A}{m}$. Rewriting the given line as $y=-ax+c$ and comparing:

$$m=-a, \qquad c=\frac{\sqrt2}{m} = \frac{\sqrt2}{-a} = -\frac{\sqrt2}{a}$$

Step 4 — Eliminate a using this relation

From $c=-\dfrac{\sqrt2}{a}$: $ac=-\sqrt2$, so squaring gives $a^2c^2=2$ …(ii)

Step 5 — Substitute a² from equation (i) into equation (ii)

From (i): $a^2=c^2-1$. Substituting into (ii):

$$(c^2-1)c^2 = 2 \implies c^4-c^2-2=0$$

Step 6 — Solve the quadratic in c²

Let $u=c^2$:

$$u^2-u-2=0 \implies (u-2)(u+1)=0 \implies u=2 \text{ or } u=-1$$

Since $u=c^2\geq0$, reject $u=-1$. So $c^2=2$.

Step 7 — Take the square root

$$|c| = \sqrt2$$

Answer

Common mistakes

• Confusing which curve’s parameter is $A$ and mismatching it against the circle’s radius. The parabola’s own constant, here $\sqrt2$ (from $y^2=4\sqrt2x$), is unrelated to the circle’s radius ($1$, from $x^2+y^2=1$) — keeping the two tangency conditions and their respective constants clearly separate avoids cross-contaminating the equations.
• Forgetting to reject $c^2=-1$. Since $c$ is a real number, $c^2$ can never be negative — this root must be discarded on sight, without needing further justification.

Practise next

• If the line $ax+y=c$ touches both the curves $x^2+y^2=4$ and $y^2=8x$, find $|c|$, using the same combined tangency-condition approach.