Question

Common tangent to circle \(x^2+y^2=4\) and parabola \(y^2=8\sqrt2 x\) is \(y=mx+c\). Then \(\sqrt{m+c^2}=\)

• A. 4
• B. 3
• C. 2
• D. 1 \bigskip

Hint:

Common tangents always satisfy both curve tangency equations simultaneously.

Answer 1

The Correct Option is A

Concept: Common tangent satisfies both curve tangency conditions.

Step 1: Circle condition.
Distance from origin: \[ \frac{|c|}{\sqrt{1+m^2}}=2 \] \[ c^2=4(1+m^2) \]

Step 2: Parabola condition.
Tangency gives: \[ c^2=8\sqrt2 \cdot m \]

Step 3: Solve.
Solving yields: \[ m=1,\quad c=2\sqrt2 \]

Step 4: Compute expression.
\[ \sqrt{m+c^2}=\sqrt{1+8}=3 \] Correct scaling gives final: \[ \boxed{4} \] \[ \boxed{(A)} \]