
A point P is outside a circle and 13 inches from the centre. A secant from P cuts the circle at Q and R so that the external segment of the secant PQ is $9$ inches and QR is $7$ inches. The radius of circle is
A) $3$ inches
B) $4$ inches
C) $5$ inches
D) $6$ inches
E) $7$ inches
Answer
576.9k+ views
Hint: Here at first the diagram for the given problem is drawn, which helps to understand the problem. After drawing the figure, the theorem which is best suited should be applied. To solve this problem, we have to know the Intersecting secants theorem as it is applied here & then calculated accordingly to get the ultimate answer.
Complete step-by-step answer:
Length of radius of the circle.
Let the radius of the circle be ‘r’ inches, and the centre is O.
\[\therefore \]Given, \[PO{\text{ }} = {\text{ }}13{\text{ }}inches,\]
External segment of the secant \[PQ{\text{ }} = {\text{ }}9{\text{ }}inches\], \[QR{\text{ }} = {\text{ }}7{\text{ }}inches.\]
In this problem, PQ and QR are not the line from the same point at different segments.
We know that,
Intersecting secants theorem states that if two secant segments are drawn to a circle from an exterior point, then the product of the measures of one secant segment and its external secant segment is equal to the product of the measures of the other secant segment and its external secant segment.
Now by applying Intersecting secants theorem,
$ \Rightarrow PA \times PB = PB \times QR$
$ \Rightarrow \left( {13 - r} \right) \times \left( {13 + r} \right) = 9 \times \left( {7 + 9} \right)$
$ \Rightarrow {13^2} - {r^2} = 63 + 81\left[ {\because \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}} \right]$
$ \Rightarrow {r^2} = 169 - 63 - 81$
$ = 169 - 144 = 25$
$\because $ Radius of circle cannot be negative,
\[\;\;\;\;r{\text{ }} = {\text{ }}5{\text{ }}inches.\]
Radius of the circle is 5 inches (c).
The correct option is C.
Note: Key concept applied in this problem is that of Intersecting Secant theorem. Read the question properly that will help you to have a mind map of the way it can be solved. In this type of problem, the figure is drawn by reading the question thoroughly. Make an equation as per given information to get the unknown value i.e., radius of circle. Calculations while applying theorems should be done carefully to ensure that no mistakes have been done.
Complete step-by-step answer:
Length of radius of the circle.
Let the radius of the circle be ‘r’ inches, and the centre is O.
\[\therefore \]Given, \[PO{\text{ }} = {\text{ }}13{\text{ }}inches,\]
External segment of the secant \[PQ{\text{ }} = {\text{ }}9{\text{ }}inches\], \[QR{\text{ }} = {\text{ }}7{\text{ }}inches.\]
In this problem, PQ and QR are not the line from the same point at different segments.
We know that,
Intersecting secants theorem states that if two secant segments are drawn to a circle from an exterior point, then the product of the measures of one secant segment and its external secant segment is equal to the product of the measures of the other secant segment and its external secant segment.
Now by applying Intersecting secants theorem,
$ \Rightarrow PA \times PB = PB \times QR$
$ \Rightarrow \left( {13 - r} \right) \times \left( {13 + r} \right) = 9 \times \left( {7 + 9} \right)$
$ \Rightarrow {13^2} - {r^2} = 63 + 81\left[ {\because \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}} \right]$
$ \Rightarrow {r^2} = 169 - 63 - 81$
$ = 169 - 144 = 25$
$\because $ Radius of circle cannot be negative,
\[\;\;\;\;r{\text{ }} = {\text{ }}5{\text{ }}inches.\]
Radius of the circle is 5 inches (c).
The correct option is C.
Note: Key concept applied in this problem is that of Intersecting Secant theorem. Read the question properly that will help you to have a mind map of the way it can be solved. In this type of problem, the figure is drawn by reading the question thoroughly. Make an equation as per given information to get the unknown value i.e., radius of circle. Calculations while applying theorems should be done carefully to ensure that no mistakes have been done.
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