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Thread: 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 $\displaystyle \triangle PCD$ 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 $\displaystyle C$ meets $\displaystyle AB$ at $\displaystyle D$.

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

    $\displaystyle \Rightarrow \triangle$'s $\displaystyle BDC, ADC$ are isosceles

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

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

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


    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 $\displaystyle A$ and $\displaystyle B$ 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 $\displaystyle DB = DC = DA = 4$ cm. Hence $\displaystyle AB = 8$ cm.

    Grandad
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