The locus of a point from which the lengths of the tangents to the circles x2+y2=4 and 2(x2+y2)−10x+3y−2=0 are equal to
A
a straight line inclined at Π4 with the line joining the centres of the circles
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B
a circle
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C
an ellipse
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D
a straight line perpendicular to the line joint the centres of the circles
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Solution
The correct option is D a straight line perpendicular to the line joint the centres of the circles Given circles may be written as, x2+y2=22 and (x−52)2+(y+34)2=(5√54)2 ⇒r1=2,C1=(0,0),r2=5√54,C2=(52,−34) And let the point from which tangent is drawn to both the circle be P(h,k) Clearly △PT1C1 and ΔPT2C2 are right angled at T1 and T2 respectively. ⇒(PT1)2=(PC1)2−r21=h2+k2−4 and (PT2)2=(PC2)2−r22=(h−52)2+(k+34)2−(5√54)2=h2+k2−5h+32k−1 Now using given condition PT1=PT2
⇒(PT1)2=(PT2)2 ⇒5h−32k=3
⇒10h−3k=6 Hence, required locus of P(h,k) is 10x−3y=6. Clearly this a straight line perpendicular to line joining C1C2.