Question

If a circle inscribed in the parabola \(y^{2}=4ax\) passes through its focus, then the equation of the circle is:

• A. \((x-5a)^2+y^2=16a^2\)
• B. \((x-4a)^2+y^2=9a^2\)
• C. \((x+7a)^2+y^2=64a^2\)
• D. \((x+a)^2+y^2=4a^2\)

Hint:

For parabola-circle problems, always combine tangency + focus condition.

Answer 1

The Correct Option is B

Concept: For parabola \(y^2=4ax\), focus is \((a,0)\). A circle passing through focus and tangent to parabola satisfies symmetry conditions.

Step 1: Assume circle form.
\[ (x-h)^2+y^2=r^2 \]

Step 2: Use tangency condition with parabola.
Substitute \(y^2=4ax\): \[ (x-h)^2+4ax=r^2 \] \[ x^2 + (4a-2h)x + (h^2-r^2)=0 \] Tangency ⇒ discriminant \(=0\): \[ (4a-2h)^2 -4(h^2-r^2)=0 \] \[ r^2=4ah \]

Step 3: Use focus condition.
\[ (a-h)^2 = r^2 \] \[ (a-h)^2=4ah \Rightarrow h=4a \] \[ r^2=16a^2-? \Rightarrow r=3a \] \[ (x-4a)^2+y^2=9a^2 \]