JEE MAIN - Mathematics (2021 - 26th August Evening Shift - No. 15)
A circle C touches the line x = 2y at the point (2, 1) and intersects the circle
C1 : x2 + y2 + 2y $$-$$ 5 = 0 at two points P and Q such that PQ is a diameter of C1. Then the diameter of C is :
C1 : x2 + y2 + 2y $$-$$ 5 = 0 at two points P and Q such that PQ is a diameter of C1. Then the diameter of C is :
$$7\sqrt 5 $$
15
$$\sqrt {285} $$
$$4\sqrt {15} $$
Explanation
(x $$-$$ 2)2 + (y $$-$$ 1)2 + $$\lambda$$(x $$-$$ 2y) = 0
C : x2 + y2 + x($$\lambda$$ $$-$$ 4) + y($$-$$2 $$-$$2$$\lambda$$) + 5 = 0
C1 : x2 + y2 + 2y $$-$$ 5 = 0
S1 $$-$$ S2 = 0 (Equation of PQ)
($$\lambda$$ $$-$$ 4)x $$-$$ (2$$\lambda$$ + 4)y + 10 = 0 Passes through (0, $$-$$1)
$$\Rightarrow$$ $$\lambda$$ = $$-$$7
C : x2 + y2 $$-$$ 11x + 12y + 5 = 0
= $${{\sqrt {245} } \over 4}$$
Diameter = $$7\sqrt 5 $$
C : x2 + y2 + x($$\lambda$$ $$-$$ 4) + y($$-$$2 $$-$$2$$\lambda$$) + 5 = 0
C1 : x2 + y2 + 2y $$-$$ 5 = 0
S1 $$-$$ S2 = 0 (Equation of PQ)
($$\lambda$$ $$-$$ 4)x $$-$$ (2$$\lambda$$ + 4)y + 10 = 0 Passes through (0, $$-$$1)
$$\Rightarrow$$ $$\lambda$$ = $$-$$7
C : x2 + y2 $$-$$ 11x + 12y + 5 = 0
= $${{\sqrt {245} } \over 4}$$
Diameter = $$7\sqrt 5 $$
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