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Minimum Circle Touching y=x²+2x+4 and x=y²+2y+4 | JEE

JEE Maths question with a full step-by-step solution.

Question
The equation of the circle of minimum radius which touches both the parabolas y=x2+2x+4y = x^2 + 2x + 4 and x=y2+2y+4x = y^2 + 2y + 4 is:
A4x2+4y2−11x−11y−31=04x^2 + 4y^2 - 11x - 11y - 31 = 0
B4x2+4y2−11x−11y−13=04x^2 + 4y^2 - 11x - 11y - 13 = 0correct
C4x2+4y2−11x−11y+11=04x^2 + 4y^2 - 11x - 11y + 11 = 0
D4x2+4y2−11x−11y−6=04x^2 + 4y^2 - 11x - 11y - 6 = 0
Solution
Step 1: The parabolas are mirror images in y=xy = x. On y=x2+2x+4y = x^2 + 2x + 4 the point nearest to y=xy = x has tangent slope 11:
2x+2=1⇒(−12, 134),2x + 2 = 1 \Rightarrow \left(-\frac{1}{2},\ \frac{13}{4}\right),
 and by symmetry the other parabola gives (134, −12)\left(\dfrac{13}{4},\ -\dfrac{1}{2}\right). These are the ends of the smallest diameter. Step 2: The circle on this diameter is
(x+12)(x−134)+(y−134)(y+12)=0⇒x2+y2−114x−114y−134=0,\left(x + \frac{1}{2}\right)\left(x - \frac{13}{4}\right) + \left(y - \frac{13}{4}\right)\left(y + \frac{1}{2}\right) = 0 \Rightarrow x^2 + y^2 - \frac{11}{4}x - \frac{11}{4}y - \frac{13}{4} = 0,
 i.e. 4x2+4y2−11x−11y−13=0.4x^2 + 4y^2 - 11x - 11y - 13 = 0. Correct answer: (2)
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