# Ex.11.2 Q3 Constructions Solution - NCERT Maths Class 10

## Question

Draw a circle of radius \(3 \,\rm{cm}\). Take two points \(P\) and \(Q\) on one of its extended diameter each at a distance of \(7 \,\rm{cm}\) from its center. Draw tangents to the circle from these two points \(P\) and \(Q\).

## Text Solution

**Steps:**

**Steps of construction:**** **

**(i)** Draw a circle with \(O\) as centre and radius is \(3 \,\rm{cm}\).

**(ii)** Draw a diameter of it extend both the sides and take points \(P\)*,* \(Q\) on the diameter such that \(\begin{align}\rm{OP}=\rm{OQ}=7 \rm{cm}\end{align}\)

**(iii)** Draw the perpendicular bisectors of \(OP\) and \(OQ\) to intersect \(OP\) and \(OQ\) at \(M\) and \(N\) respectively.

**(iv)** With \(M\) as centre and \(OM\) as radius draw a circle to cut the given circle at \(A\) and \(C\). With \(N\) as centre and \(ON\) as radius draw a circle to cut the given circle at \(B\) and \(D\).

**(v)** Join \(PA\), \(PC\) \(QB\), \(QD\)

\(PA\),\(PC\) and \(QB\),\(QD\) are the required tangents from \(P\) and \(Q\) respectively.

**Proof:**

\(\angle {{PAO}} =\angle {{QBO}}=90^\circ \) (Angle in a semi - circle)

\(\therefore {PA} \perp {AO},\; {QB} \perp {BO}\)

Since \(OA\) and \(OB\) are the radii of the given circle, \(PA\) and QB are its tangents at \(A\) and \(B\) respectively.

In right angle triangle \(PAO\) and \(QBO\)

\(OP = OQ = 7 \,\rm{cm }\) (By construction)

\(OA = OB = 3\,\rm{cm}\) (radius of the given circle)

\[\begin{align} {{P}}{{{A}}^{{2}}}&={{ (OP}}{{{)}}^{{2}}}{{ -(OA}}{{{)}}^{{2}}}\,\,\,\,\,\,\,\\ &= {{ (7}}{{{)}}^{{2}}}{{ - (3}}{{{)}}^{{2}}}\\ &= {{ 49 - 9}}\\ &={{ 40}}\\ {{PA }}&=\sqrt {{{40}}} \\ &={{ 6}}{{.3}}\,\,{\rm{(approx)}} \end{align}\]

And

\[\begin{align} {{Q}}{{{B}}^{{2}}}&={{(OQ}}{{{)}}^{{2}}}{{ - (OB}}{{{)}}^{{2}}}\,\,\,\,\,\,\,\\ &={{(7}}{{{)}}^{{2}}}{{ - (3}}{{{)}}^{{2}}}\\ &={{ 49 - 9}}\\ &={{ 40}}\\ {{QB }}&=\sqrt {{{40}}} \\ &={{ 6}}{{.3}}\,\,{\rm{(approx)}} \end{align}\]