Question

two non intersecting circles of equal radii have their centres A and B if point P lies on the perpendicular bisector of ab show that the length of tangents PQ and PR drawn from p to the circle with centres A and B are equal

Answer 1

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Answer 2

Answer:

Two non intersecting circles having centers A and B , and radii equal to r.

Join A and B.

Draw perpendicular bisector of AB.Let perpendicular bisector of segment AB cuts line AB at point M, and point P lies on  the perpendicular bisector of AB.

Then, AP=BP, as perpendicular bisector of a segment is equidistant from two ends of line segment.

Draw tangents PQ of circle having center A, and another tangent PR of circle,  having center B.

∠AQP=∠BRP=90°→→[Line from center of circle to the point of contact of tangent makes an angle of 90°, with the tangent of circle.]

In ΔAQP and ΔBRP

AQ=BR→→[Radii of two circles are equal]

∠AQP=∠BRP=90°→→→[Proved above]

AP=BP→→[Proved above]

∠AQP ≅∠BRP→→[By R HS, it states that if in two triangles hypotenuse and one side of triangle is equal to Hypotenuse and other side of triangle , it means two triangles are congruent]

So, PQ=PR→→[CPCT]