Question

In an online test, Ishita comes across the statement - If a tangent is drawn to a circle from an external point, then the square of length of tangent drawn is equal to difference of squares of distance of the tangent from the centre of circle and radius of the circle.Help Ishita, in answering the following questions based on the above statement.
(i) If $AB$ is a tangent to a circle with centre $O$ at $B$ such that $AB=10cm$ and $OB=5cm$, then $OA=$
(a) $3\sqrt{5}cm$
(b) $5\sqrt{5}cm$
(c) $4\sqrt{5}cm$
(d) $6\sqrt{5}cm$

(ii) In the adjoining figure, radius of the circle is

(a) $8cm$
(b) $7cm$
(c) $9cm$
(d) $10cm$

(iii)In the adjoining figure, length of tangent $AP$ is

(a) $12cm$
(b) $24cm$
(c) $30cm$
(d) None of these

(iv) $PT$ is a tangent to a circle with centre $O$ and diameter $=40cm$. If $PT=21cm$, then $OP=$
(a) $33cm$
(b) $29cm$
(c) $37cm$
(d) None of these

(v) In the adjoining figure, the length of the tangent is

(a) $15cm$
(b) $9cm$
(c) $8cm$
(d) $10cm$

Answer 1

(i) (b): $O{A}^2=A{B}^2+O{B}^2$
$$\Rightarrow OA=\sqrt{{10}^2+5^2}=5\sqrt{5}cm$$(ii) (a): $OA=\sqrt{O{P}^2-A{P}^2}$
(Given)
$$=\sqrt{{17}^2-{15}^2}=\sqrt{64}=8cm$$
(iii) (b): Length of tangent, $AP=\sqrt{O{P}^2-O{A}^2}$ (Given)
$$=\sqrt{{25}^2-7^2}=\sqrt{576}=24cm$$


(iv) (b):
Since, $OP=\sqrt{(PT{)}^2+(OT{)}^2}=\sqrt{{21}^2+{20}^2}=29cm$


(v) (a): Since, $O{P}^2+P{Q}^2=O{Q}^2$ $\Rightarrow 8^2+x^2=(x+2{)}^2\Rightarrow 64=4x+4\Rightarrow x=15cm$ So, length of tangent, $PQ=15cm$