# Ex.11.2 Q5 Constructions Solution - NCERT Maths Class 10

## Question

Draw a line segment \(AB\) of length \(8 \,\rm{cm}\). Taking \(A\) as centre, draw a circle of radius \(4 \,\rm{cm}\) and taking \(B\) as center, draw another circle of radius \(3 \,\rm{cm}\). Construct tangent to each circle from the center of the other circle.

## Text Solution

**Steps:**

**Steps of construction:**

**(i)** Draw \(\begin{align}\rm{AB}=8\, \rm{cm}\end{align}\).With \(A\) and \(B\) as centers \(4 \,\rm{cm}\) and \(3 \,\rm{cm}\) as radius respectively draw two circles.

**(ii)** Draw the perpendicular bisector of \(AB\), intersecting \(AB\) at \(O\).

**(iii)** With \(O\) as center and *\(OA\)* as radius draw a circle which intersects the two circles at \(P\), \(Q\), \(R\) and \(S\)*.*

**(iv)** Join \(BP\), \(BQ\), \(AR\) and \(AS\).

**(v)** \(BP\) and \(BQ\) are the tangents from \(B\) to the circle with center \(A\). \(AR\) and \(AS\) are the tangents from \(A\) to the circle with center \(B\).

** ****Proof:**

**\(\angle {APB}=\angle {AQB}=90^{\circ}\)** (Angle in a semi-circle)

\(\therefore \rm{AP} \perp \rm{PB}\) and \({AQ} \perp {QB}\)

Therefore, \(BP\) and \(BQ\) are the tangents to the circle with center \(A\)*.*

Similarly, \(AR\) and \(AS\) are the tangents to the circle with center \(B\).