# Math Help - Circles problem.

1. ## 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.

2. ## Tangent to a circle

Hello aman
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.

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.