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← JEE Main 2022 · 25 Jul · Shift 2
MathematicsInteger+4 / -1Question 28 of 90

JEE Main 2022 · 25 Jul · Shift 2, Mathematics question 28

If the circles x2+y2+6x+8y+16=0x^2+y^2+6x+8y+16=0 and x2+y2+2(3−3)x+2(4−6)y=k+63+86x^2+y^2+2\left(3-\sqrt3\right)x+2\left(4-\sqrt6\right)y=k+6\sqrt3+8\sqrt6, k>0k>0, touch internally at the point P(α,β)P(\alpha,\beta), then (α+3)2+(β+6)2\left(\alpha+\sqrt3\right)^2+\left(\beta+\sqrt6\right)^2 is equal to ______.

Answer and worked solution

Official answer: 25

Circles. C1=(−3,−4)C_1=(-3,-4), r1=3r_1=3; C2=(−3+3,−4+6)C_2=(-3+\sqrt3,-4+\sqrt6), ∣C1C2∣=3|C_1C_2|=3, so r2=6r_2=6.

Contact P=2C1−C2=(−3−3,−4−6)P=2C_1-C_2=(-3-\sqrt3,-4-\sqrt6); the value is 9+16=259+16=25.

Answer: 25