# Circles problem.

• Aug 1st 2009, 11:26 PM
aman
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.
• Aug 2nd 2009, 09:55 AM
Tangent to a circle
Hello aman
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Originally Posted by aman
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.

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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.

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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.

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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\$

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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.