Question

In the given figure, $BDC$ is a tangent to the given circle at point $D$ such that $BD=30$ cm and

$CD$ = $7$ cm. The other tangents $BE$ and $CF$ are dawn respectively from $B$ and $C$ to the circle and meet when produced at $A$ making $BAC$ a right angle triangle.

Calculate (i) $AF$ (ii) radius of the circle.

Answer 1

(i)

AB, BC and AC are tangents to the circle at E, D and F.

$BD=30cm,DC=7cm,\angle BAC=90$

From the theorem stated,

$BE=BD=30cm$

Also $FC=DC=7cm$

Let $AE=AF=x$ …. (1)

Then $AB=BE+AE=(30+x)$

$AC=AF+FC=(7+x)$

$BC=BD+DC=30+7=37cm$

Consider right trianlge ABC, by Pythagoras theorem we have

$B{C}^2=A{B}^2+A{C}^2$

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

$1369=900+60x+x^2+49+14x+x^2$

$2x^2+74x+949–1369=0$

$2x^2+74x–420=0$

$x^2+37x–210=0$

$x^2+42x–5x–210=0$

$x(x+42)–5(x+42)=0$

$(x–5)(x+42)=0$

$(x–5)=0or(x+42)=0$

$x=5orx=–42$

$x=5$[Since x cannot be negative]

$\therefore AF=5cm$[From (1)]

Therefore$AB=30+x=30+5=35cm$

(ii)

$AC=7+x=7+5=12cm$

Let ‘O’ be the centre of the circle and ‘r’ the radius of the circle.

Join point O, F; points O, D and points O, E.

From the figure,

$\frac{1}{2}\times AC\times AB=\frac{1}{2}\times AB\times OE+\frac{1}{2}\times BC\times OD+\frac{1}{2}\times AC\times OC$

$AC\times AB=AB\times OE+BC\times OD+AC\times OC$

$12\times 35=35\times r+37\times r+12\times r$

$420=84r$

$\therefore r=5$

Thus the radius of the circle is 5 cm.