If the circles x² + y² + 6x + 8y + 16 = 0 and x² + y² + 2(3 - 3)x + 2(4 - 6)y = k + 6 3 + 8 6, k > 0, touch internally at the point P(α,β), then (α + 3)² + (β + 6)² is equal to _ _ _ _ _ _ _ _.

Coordinate Geometry: JEE Main 2022 Mathematics question

JEE Main 2022 · 25 Jul, Evening shift · Q29

JEE Main 202225 Jul, Evening shiftMathematicsMediumCoordinate Geometry478× since 2002

If the circles

{x^2} + {y^2} + 6x + 8y + 16 = 0

and

{x^2} + {y^2} + 2\left( {3 - \sqrt 3 } \right)x + 2\left( {4 - \sqrt 6 } \right)y = k + 6\sqrt 3 + 8\sqrt 6

k > 0

, touch internally at the point

P(\alpha ,\beta )

, then

{\left( {\alpha + \sqrt 3 } \right)^2} + {\left( {\beta + \sqrt 6 } \right)^2}

is equal to ________________.

Your answerType your answer. Integers and decimals are accepted.