Circumscribed Quadrilateral:
A quadrilateral \(ABCD\) is said to circumscribe a circle if all four of its sides touch the circle tangentially.
The circle inside is called a incircle, and the points where the sides touch the circle are called points of contact.
Theorem on circles and tangents:
Theorem \(1\):
Statement:
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Proof:
Consider a circle with centre \(O\).
Let \(AB\) be the tangent to the circle at the point \(P\).
To prove:
The line \(OP\) is perpendicular to \(AB\).
Proof:
Take a point \(Q\) other than \(P\) on the tangent \(AB\) and join \(OQ\).
Here, \(Q\) must lie outside the circle.
Thus, \(OQ\) is longer than \(OP\).
That is \(OQ>OP\) at every point on \(AB\) except at \(P\).
Therefore, the point \(P\) is at the shortest distance from the centre \(O\).
We know that:
Out of all the line segments drawn from the point to points of a line not passing through the point, the smallest is the perpendicular to the line.
By the theorem, \(OP\) is perpendicular to \(AB\).
Hence, the proof.
Theorem \(2\):
Statement:
The lengths of tangents drawn from an exterior point to a circle are equal.
Proof:
Consider a circle with centre \(O\).
Let \(PA\) and \(PB\) be the two tangents drawn from the external point \(P\) to the circle.
Construction:
Join \(OA\), \(OB\) and \(OP\).
To prove:
The tangent \(PA=\) The tangent \(PB\).
Proof:
By theorem \(1\), we have:
The tangent at any point of a circle is perpendicualr to the radius through the point of contact.
\(OB\perp PB\) and \(OA\perp PA\)
Here, \(OA\) and \(OB\) are radius. Hence, they are equal.
The side \(OP\) is a common side to the triangles \(AOP\) and \(BOP\).
Therefore, by \(RHS\) rule, if the length of the hypotenuse and one side of one triangle is equal to the length of the hypotenuse and corresponding side of the other triangle, then the two triangles are congruent.
The triangles \(AOP\) and \(BOP\) are congruent.
We know that the corresponding parts of the congruent triangles are equal.
Therefore, \(PA=PB\).
Example:
A quadrilateral is circumscribed about a circle. If \(AB = 8\ cm\), \(BC = 11\ cm\) and \(CD = 7\ cm\). Find \(AD\).
Solution:
Given \(ABCD\) is a quadrilateral.
Let \(P\), \(Q\), \(R\) and \(S\) be the point of contacts.
We know that, the lengths of tangent from an external point are equal.
\(AP = AQ\), \(BP = BS\), \(CR = CS\) and \(DR = DQ\)
Adding \(LHS\) and \(RHS\) individually we get,
\(AP + BP + CR + DR = AQ + DQ + BS + CS\)
\(AB + CD = AD + BC\)
\(8 + 7 = AD + 11\)
\(15 - 11 = AD\)
\(AD = 4\ cm\)
Therefore, the length of \(AD\) is \(4 \ cm\)
Important!
If a quadrilateral is circumscribed about a circle, then the sum of the lengths of opposite sides are equal, \(AB + CD = BC + AD\).