In the given figure, ^@ AB ^@ is a chord of length ^@ 13 \space cm ^@ of a circle with center ^@ O ^@ and radius ^@ 9 \space cm ^@. The tangents at ^@ A ^@ and ^@ B ^@ intersect at ^@ P ^@. Find the length of ^@ PA ^@.

O B A P


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Answer:

^@ 9.4 \space cm ^@

Step by Step Explanation:
  1. Given:
    Length of chord ^@ AB ^@ is ^@ 13 \space cm ^@.
    The radius of the circle is ^@ 9 cm ^@.
    The tangents at ^@ A ^@ and ^@ B ^@ intersect at ^@ P ^@.
  2. Here, we have to find the length of ^@ PA ^@.

    Now, join ^@O^@ to ^@P^@ such that ^@OP^@ intersects ^@AB^@ at ^@M^@.

    Let ^@ PA = x \space cm ^@ and ^@ PM = y \space cm^@.
    x cm 13 cm 9 cm O B A P y cm M
  3. The tangents from an external point are equal in length. @^ \implies PA = PB @^ Also, two tangents to a circle from an external point are equally inclined to the line segment joining the center to that point. @^ \begin{aligned} \implies & OP \text{ is the bisector of } \angle APB. \\ \implies & \angle APM = \angle MPB \end{aligned} @^ Also, @^ \begin{aligned} PO = PO && \text{[Common]} \end{aligned} @^ By SAS Congruence Criterion, we conclude @^ \begin{aligned} \triangle APM \cong \triangle MPB \end{aligned} @^
  4. As corresponding parts of congruent triangles are equal^@(CPCT)^@, we have @^ \angle PMA = \angle PMB \text{ and } AM = BM @^ Also, @^ \begin{aligned} & \angle PMA + \angle PMB = 180^\circ && \text{ [Angles on a straight line] } \\ \implies & 2 \angle PMA = 180^\circ && \text{ [As, } \angle PMA = \angle PMB] \\ \implies & \angle PMA = \angle PMB = 90^\circ \end{aligned} @^
  5. Now, we can say that ^@ OP ^@ is the right bisector of ^@AB^@.

    Thus, ^@OP \perp AB ^@ and ^@OP ^@ bisects ^@ AB ^@ at ^@ M ^@.

    Therefore, ^@ AM = BM = \dfrac { 13 } { 2 } \space cm = 6.5 \space cm ^@.
  6. Now, @^ \begin{aligned} & \angle PMA = 90^\circ \implies \angle AMO = 90^\circ && \text{ [Angles on a straight line.] } \\ \implies & \triangle AMO \text{ is a right-angled triangle.} \end{aligned} @^
    x cm 9 cm 6.5 cm 6.5 cm O B A P y cm M

    In right ^@ \triangle AMO ^@, we have @^ \begin{aligned} OA = 9 \space cm && \text{[Radius]} \\ AM = 6.5 \space cm && \text{[From step 5]} \end{aligned} @^ Therefore, using pythagoras theorem, we have @^ \begin{aligned} OM^2 & = AM^2 + OM^2 \\ OM & = \sqrt{(9)^2 - (6.5)^2} \space cm \\ & = \sqrt{ 38.75 } \space cm \\ & = 6.22 \space cm \end{aligned} @^
  7. Also, ^@ \triangle APM ^@ is a right angled triangle.
    Using pythagoras theorem, we have @^ \begin{aligned} & PA^2 = PM^2 + AM^2 \\ \implies & x^2 = y^2 + (6.5)^2 \\ \implies & x^2 = y^2 + 42.25 && \ldots \text{(i)} \end{aligned} @^
  8. In right ^@ \triangle PAO ^@, using pythagoras theorem @^ \begin{aligned} & OP^2 = PA^2 + OA^2 && [\angle PAO = 90^\circ \text{ as AO is the } \space \\ & && \text { radius at the point of contact.] } \\ \implies & (PM + OM)^2 = x^2 + (9)^2 && \text{[As, } OP = PM + OM ] \\ \implies & PM^2 + OM^2 + 2 \times PM \times OM = x^2 + 81 \\ \implies & y^2 + 38.75 + 2y \times 6.22 = x^2 + 81 \\ \implies & y^2 + 38.75 + 2y \times 6.22 = y^2 + 42.25 + 81 && \text{[By using } eq \text{(i)} ] \\ \implies & 12.44 y = 84.5 \\ \implies & y = 6.79 \space cm \end{aligned} @^
  9. Now, substituting the value of ^@ y ^@ in ^@ \text{(i)}^@, we get @^ \begin{aligned} & x^2 = (6.79)^2 + 42.25 = 46.14 + 42.25 = 88.39 \\ \implies & x = \sqrt { 88.39 } = 9.4 \end{aligned} @^
    Thus, ^@ PA = 9.4 \space cm.^@

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