Question

B D C

is a tangent to the given circle at point

D

such that

B D = 30 cm

and

C D = 7 cm

. The other tangents

B E

and

C F

are drawn respectively from

B

and

C

to the circle and meet when produced at

A

making

B A C

a right-angled triangle. Calculate radius of the circle.

Accepted Answer

Let us join the point of contact $E$ with the centre of the circle say $D$ . Also, let us join the point of contact $B D = 30 cm$ with the centre of the circle $O$ . Now we have a quadrilateral $A E O F$ .

$A B, B C$ and $A C$ are tangents to the circle at $E, D$ and $F$ .

$O$ is the centre of the circle and radii $O E = O F$ .

As radius is perpendicular to the tangent at the point of contact, $∠ O E A = ∠ O F A = 90 °$ .

As $∠ B A C = ∠ E A F = 90 °$ , the quadrilateral $O E A F$ is a square.

Therefore, $O E = E A = A F = F O = x$ .

$B D = 30 cm$ and $D C = 7 cm$ and $∠ B A C = 90 °$

$B E = B D = 30 cm$ (Length of tangents from an external point are equal)

Also, $F C = D C = 7 cm$

Let, $A E = A F = x \dots . (1)$

Then, $A B = B E + A E = (30 + x)$

$A C = A F + F C = (7 + x)$

$B C = B D + D C = 30 + 7 = 37 cm$

Consider right $△ A B C$ , by Pythagoras theorem we have, $B C^{2} = A B^{2} + A C^{2}$ .

$\Rightarrow (37)^{2} = (30 + x)^{2} + (7 + x)^{2}$

$\Rightarrow 1369 = 900 + 60 x + x^{2} + 49 + 14 x + x^{2}$

$\Rightarrow 2 x^{2} + 74 x + 949 - 1369 = 0$

$\Rightarrow 2 x^{2} + 74 x - 420 = 0$

$\Rightarrow x^{2} + 37 x - 210 = 0$

Solving the equation $x^{2} + 37 x - 210 = 0$ by splitting the middle term,

$x^{2} + 42 x - 5 x - 210 = 0$

$\Rightarrow x (x + 42) - 5 (x + 42) = 0$

$\Rightarrow (x - 5) (x + 42) = 0$

$\Rightarrow (x - 5) = 0$ or $(x + 42) = 0$

$\Rightarrow x = 5$ or $x = - 42$

$\Rightarrow x = 5 [since x cannot be negative]$

$∴ A F = 5 cm [From (1)]$

Therefore, $A B = 30 + x = 30 + 5 = 35 cm$ .

Therefore, radius of the circle is $5 cm$ .

EXERCISE 8.2

R. D. Sharma Mathematics Solutions for EXERCISE 8.2

Questions from EXERCISE 8.2 with Hints & Solutions

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