Ex.10.2 Q10 Circles Solution - NCERT Maths Class 10

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Question

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the center.

Text Solution

What is Known?

  • Let us consider a circle centered at point \({O.}\)
  • Let \({P}\) be an external point from which two tangents \({PA}\) and \({PB}\) are drawn to the circle which are touching the circle at point \({A}\) and \({B}\) respectively
  • \({AB}\) is the line segment, joining point of contacts \({A}\) and \({B}\) together such that it subtends \(\angle {AOB}\) at center \({O}\) of the circle.

To prove:

The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the point of contact at the center.

i.e. \(\angle {APB}\) is supplementary to \(\angle {AOB}\)

Reasoning:

According to Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

\(\begin{align}\therefore  \angle {OAP} = \angle {OBP} = 90^{ \circ } \ldots \end{align}\)  Equation (i)

Steps :

In a quadrilateral, sum of 4 angles is \({360^ \circ }\)

\(\therefore\) In \({OAPB}\),

\[\begin{align}\begin{bmatrix}\angle {OAP} + \angle {APB} + \\\angle {PBO} + \angle {BOA} \end{bmatrix}= 360 ^ { \circ }\end{align}\]

Using Equation (i), we can write the above equation as

\[\begin{align}   \left[ \begin{array}  & {{90}^{{}^\circ }}+\angle APB+ \\  {{90}^{{}^\circ }}+\angle BOA \\ \end{array} \right]&\!\!=\!{{360}^{^\circ }} \\   \angle APB+\angle BOA&\!\!=\!{{360}^{{}^\circ }}\!\!-\!\!{{180}^{{}^\circ }} \\\therefore \angle APB+\angle BOA&\!\!=\!\!{{180}^{{}^\circ }} \\ \end{align}\]

Where, \(\angle {APB}=\) Angle between the two tangents \({PA}\) and \({PB}\) from external point \({P}\).

\(\angle {BOA}=\) Angle subtended by the line segment joining the point of contact at the center.

Hence Proved.

  
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