Question

If a straight line \[ y=mx+c \] touches the circle \[ x^2+y^2=4 \] and parabola \[ y^2=4x \] then \(2m^2=\)

• A. \(\sqrt2+1\)
• B. \(2\)
• C. \(\frac12\)
• D. \(\sqrt2-1\)

Hint:

For parabola \(y^2=4ax\), tangent in slope form is always \(y=mx+\frac{a}{m}\).

Answer 1

The Correct Option is D

Concept: For tangent to parabola \[ y=mx+\frac1m \] For tangent to circle distance from center equals radius.

Step 1: Condition for parabola tangent.
Since tangent to parabola \[ c=\frac1m \] Thus line is \[ y=mx+\frac1m \]

Step 2: Condition for circle tangent.
Circle centered at origin radius 2. Distance from origin: \[ \frac{|c|}{\sqrt{1+m^2}}=2 \] Substitute \[ \frac{\frac1m}{\sqrt{1+m^2}}=2 \] Square both sides \[ \frac1{m^2(1+m^2)}=4 \] \[ 1=4m^2+4m^4 \] Let \[ u=m^2 \] Then \[ 4u^2+4u-1=0 \] \[ u=\frac{\sqrt2-1}{2} \] Thus \[ 2m^2=\sqrt2-1 \] Hence \[ \boxed{\sqrt2-1} \]