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Result \(1\):
A tangent at any point on a circle and the radius through the point are perpendicular to each other.
Explanation:
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The tangent at the point \(P\) on a circle and the radius through the point \(P\) are perpendicular.
That is, the radius \(OP\) makes an angle \(90^{\circ}\) with the tangent \(AB\) at the point \(P\).
Example:
In the above given figure if \(OP\) \(=\) \(3\) \(cm\) and \(PQ\) \(=\) \(4\) \(cm\), find the length of \(OQ\).
Solution:
By the result, \(\angle OPQ\) \(=\) \(90^{\circ}\).
So, \(OPQ\) is a right-angled triangle.
By the Pythagoras theorem, we have:
In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
\(OQ^2\) \(=\) \(OP^2\) \(+\) \(PQ^2\)
\(OQ^2\) \(=\) \(3^2\) \(+\) \(4^2\)
\(OQ^2\) \(=\) \(9 + 16\)
\(OQ^2\) \(=\) \(25\)
\(\Rightarrow\) \(OQ\) \(=\) \(\sqrt{25}\)
\(OQ\) \(=\) \(5\)
Therefore, the measure of \(OQ\) \(=\) \(5\) \(cm\)
Result \(2\):
- No tangent can be drawn from an interior point of the circle.
- Only one tangent can be drawn at any point on a circle.
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- Two tangents can be drawn from any exterior point of a circle.
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