# Circles : Problems and Solutions

Problem (1)

Given : $\displaystyle SPT$ is the tangent to the circle at $\displaystyle P$.

$\displaystyle PQ$ and $\displaystyle RQ$ are the chords of the circle.

$\displaystyle PM\bot RQ$ and $\displaystyle RN\bot SPT$.

Prove : $\displaystyle MN\parallel PQ$

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \ \angle QPT=\angle QRP\ (\angle \ \text{between tangent and chord }\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{= }\angle \ \text{in alternate segment})\\\\\ \ \ \ \ \text{Since}\ PM\bot RQ\ \text{and}\ RN\bot SPT,\ \\\\\ \ \ \ \ \angle PMR=\angle PNR=90{}^\circ \\\\\therefore \ \ \ \angle PMR+\angle PNR=180{}^\circ \\\\\therefore \ \ \ PMRN\ \text{is cyclic}\text{.}\\\\\therefore \ \ \ \angle PNM=\angle QRP\ \ (\angle \ \text{in same arc)}\\\\\therefore \ \ \ \angle PNM=\angle QPT\\\\\ \ \ \ \ \text{Since}\ \angle PNM\ \text{and}\ \angle QPT\ \text{are alternating angles,}\\\\\ \ \ \ \ MN\parallel PQ.\ \end{array}$

Problem (2)

In the fgure, $\displaystyle AP$ is a tangent to the circle at $\displaystyle A$ and $\displaystyle AP$ is parallel to $\displaystyle BQ$. Prove that

(i) $\displaystyle ∆ABC$ is similar to $\displaystyle ∆AQB,$

(ii) $\displaystyle AB^2 = AQ × AC.$

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \ \alpha =\beta \ (\angle \ \text{between tangent and chord }\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{= }\angle \ \text{in alternate segment})\\\\\ \ \ \ \ \text{Since}\ AP\parallel BQ,\ \ \alpha =\theta .\\\\\ \ \ \ \ \text{In }\Delta ABC\ \text{and}\ \Delta AQB,\\\\\ \ \ \ \ \alpha =\theta \ \text{(proved)}\\\\\ \ \ \ \ \delta =\delta \ \ \text{(common }\angle \text{)}\\\\\therefore \ \ \ \Delta ABC\sim \Delta AQB\ \ (\text{AA corollary)}\end{array}$

$\displaystyle \therefore \ \ \ \frac{{AB}}{{AQ}}=\frac{{AC}}{{AB}}$

$\displaystyle \therefore \ \ \ A{{B}^{2}}=AQ\times AC$

Problem (3)
In the fgure, $\displaystyle O$ is the centre of the circle, $\displaystyle PQ$ is a diameter and $\displaystyle AB$ is a chord which is parallel to $\displaystyle PQ.$ $\displaystyle AQ$ and $\displaystyle OB$ intersect at $\displaystyle X$. Prove that $\displaystyle ∠BXQ = 3∠PQA$.

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \text{Since}\ AB\parallel PQ,\\\\\ \ \ \ \beta =\phi .\ \ \ \text{(}\because \text{alternating }\angle \text{)}\\\\\ \ \ \ \alpha =\theta .\ \ \ \text{(}\because \text{alternating }\angle \text{)}\end{array}$

$\displaystyle \ \ \ \ \alpha =\frac{1}{2}\phi \ \ \ \text{(}\because \text{inscribed }\angle =\frac{1}{2}\text{central }\angle \text{)}$

$\displaystyle \begin{array}{l}\therefore \ \ \phi =\beta =2\alpha =2\theta \\\\\ \ \ \ \text{In }\Delta ABX,\ \\\\\ \ \ \ \gamma =\beta +\alpha \\\\\ \ \ \ \ \ \ =2\alpha +\alpha \\\\\ \ \ \ \ \ \ =3\alpha \\\\\ \ \ \ \ \ \ =3\theta \ \ \ \ (\because \alpha =\theta )\\\\\therefore \ \ \angle BXQ=3\angle PQA\end{array}$

Problem (4)
In the fgure, $\displaystyle BD$ and $\displaystyle CE$ are tangents to the circle, of which $\displaystyle AB$ is a diameter and $\displaystyle ACD$ is a straight line. Prove that

(i) $\displaystyle \angle ABC=\angle ECD$

(ii) $\displaystyle BE = ED.$

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \text{It is obvious that}\\\\\ \ \ \beta =\gamma \ \ (\because \angle \ \text{between tangent and chord}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\angle \ \text{in alternate segment})\\\\\ \ \ \text{But, }\gamma =\theta \ \ (\because \text{vertically opposite }\angle \text{s)}\\\\\therefore \ \beta =\theta \\\\\therefore \angle ABC=\angle ECD\\\\\ \ \ \ \text{Since }AB\ \text{is a diameter, }\\\\\ \ \ \ \angle ACB=90{}^\circ \ \ (\because \angle \ \text{in semicircle)}\\\\\therefore \ \ \angle BCD=90{}^\circ \\\\\ \ \ \ \text{Furthermore, }AB\ \text{is a diameter and }BD\ \text{is a tangent,}\\\\\ \ \ \ AB\bot BD.\\\\\therefore \ \ \Delta ABD\sim \Delta ACB\sim \Delta BCD\\\\\therefore \ \ \beta =\delta \\\\\therefore \ \ \theta =\delta \\\\\therefore \ \ \Delta BCD\ \text{is an isosceles triangle with base }CD.\\\\\therefore \ \ EC=ED\\\\\ \ \ \ \text{Since}\ BE=EC,\ \ (\because \text{tangents from same exterior point)}\\\\\ \ \ \ BE=ED\end{array}$

Problem (5)
In the figure, $\displaystyle ABCD$ and $\displaystyle APRD$ are two circles intersecting at $\displaystyle A$ and $\displaystyle D$. $\displaystyle ARC$ and $\displaystyle BPD$ are straight lines. $\displaystyle DR$ produced meets $\displaystyle BC$ at $\displaystyle S$. Prove that $\displaystyle PR$ is parallel to $\displaystyle BC.$

Show/Hide Solution
In circle $\displaystyle APRD, \alpha = \phi \ \ \ (\angle \text{s}\ \text{in}\ \text{same}\ \text{arc})$

Similarly, In circle $\displaystyle ABCD, \alpha = \beta \ \ \ (\angle \text{s}\ \text{in}\ \text{same}\ \text{arc})$

$\displaystyle \therefore\ \phi= \beta$

Since $\displaystyle \phi$ and $\displaystyle \beta$ are corresponding angles,

$\displaystyle PR \parallel BC$

Problem (6)
In the figure, $\displaystyle HAE$ is the tangent to the circle at $\displaystyle H, BH = BE$ and $\displaystyle KH$ is the angle bisector of $\displaystyle ∠BHE$ and it cuts the circle at $\displaystyle D.$ Given that $\displaystyle BD$ produced meets $\displaystyle HE$ at $\displaystyle A,$ prove that

(i) $\displaystyle HD = BD,$

(ii) $\displaystyle A, D, K$ and $\displaystyle E$ are concyclic.

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \beta =\phi ,\ \ \ (\angle \ \text{between tangent and chord}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\angle \ \text{in alternate segment})\end{array}$

$\displaystyle \ \ \ \text{Since}\ KH\ \text{is}\ \text{the}\ \text{angle}\ \text{bisector}\ \text{of}\ ∠BHE,$

$\displaystyle \ \ \ \theta = \phi$

$\displaystyle \therefore \theta = \beta$

$\displaystyle \therefore HD = BD$

$\displaystyle \ \ \ \text{Since}\ BH = BE,$

$\displaystyle \ \ \ \varepsilon=\theta + \phi$

$\displaystyle \therefore \varepsilon=\theta + \beta$

$\displaystyle \ \ \ \text{In}\ \triangle BDH,$

$\displaystyle \ \ \ \alpha=\theta + \beta$

$\displaystyle \therefore \varepsilon=\alpha$

$\displaystyle \text{Since}\ \alpha +\delta =180{}^\circ ,$

$\displaystyle \ \ \ \varepsilon +\delta =180{}^\circ ,$

$\displaystyle \therefore \ A,D,K,E\ \text{are concyclic}\text{.}$