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Math Help - Circles problem.

  1. #1
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    Circles problem.

    1. If from any point on the common chord of two intersecting circles , tangents be drawn to the circles , prove that they are equal.

    2. From an external point P , tangents PA and PB are drawn to a circle with center O. If CD is the tangent to the circle at a point E and PA = 14 cm , find the perimeter of triangle PCD.

    3. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the center of the circle.

    4. Two circles touch each other externally at C and AB is a common tangent to the circles. Prove that angle ACB = 90 degrees.

    5. AB and CD are two common tangent to circles which touch each other at C. If D lies on AB such that CD = 4cm , then show that AB = 8cm.
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  2. #2
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    Tangent to a circle

    Hello aman
    Quote Originally Posted by aman View Post
    1. If from any point on the common chord of two intersecting circles , tangents be drawn to the circles , prove that they are equal.
    This can be done very easily using the secant-tangent theorem, which you'll find here: Power of a point - Wikipedia, the free encyclopedia. Using the notation given on this page, the same secant PMN cuts both circles. So if the two tangents are PT and PS, say, PT^2 = PM.PN = PS^2. Hence PT = PS.

    2. From an external point P , tangents PA and PB are drawn to a circle with center O. If CD is the tangent to the circle at a point E and PA = 14 cm , find the perimeter of triangle PCD.
    This doesn't make sense. Where are C and D? They could be anywhere, so it's impossible to find the perimeter of \triangle PCD without more information.

    3. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the center of the circle.
    This is the fundamental property of a tangent to a circle: that it is perpendicular to the radius through its point of contact.

    4. Two circles touch each other externally at C and AB is a common tangent to the circles. Prove that angle ACB = 90 degrees.
    Suppose the common tangent at C meets AB at D.

    Then DB = DC = DA (tangents from a point to a circle are equal in length).

    \Rightarrow \triangle's BDC, ADC are isosceles

    \Rightarrow \angle BCD = \angle CBD = x, say; and \angle DAC = \angle ACD = y, say.

    Then, considering the angle sum of \triangle ABC: 2x + 2y = 180^o

    \Rightarrow \angle ACB = x + y = 90^o


    5. AB and CD are two common tangent to circles which touch each other at C. If D lies on AB such that CD = 4cm , then show that AB = 8cm.
    Again, this question doesn't really make sense, as you have stated it, because you haven't said where A and B are - they could lie anywhere on the common tangent.

    But, if A and B are actually the points of contact of the tangent to the two circles, then you can simply use the fact (stated in my solution to 4 above) that the tangents from a point to a circle are equal in length. Then DB = DC = DA = 4 cm. Hence AB = 8 cm.

    Grandad
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