The line 2x=y intersects the circle C1:x2+y2=5 at P in the 1st quadrant. C2 and C3 are circles, both with radii equal to 2√5 with respective centers Q2 and Q3 lying on the y-axis. If the tangent to C1 at P touches the circles C2 and C3 then the distance Q2Q3 is
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Solution
Point of intersection of the curves C1:x2+y2=5 and 2x=y is (1,2) Equation of the tangent at (1,2) ⇒x+2y=5 Let Q2 and Q3 be (0,k) ⇒∣∣∣1×0+2×k−5√5∣∣∣=2√5 ⇒2k−5=±10⇒k=7.5,−2.5⇒Q2=(0,7.5),Q3=(0,−2.5) ⇒ Distance Q2Q3=10units