# Ncert Solutions For Class 9 Maths Ex 10.3

## Ncert Solutions For Class 9 Maths Chapter 10 Ex 10.3

Exemplar problems

1. If V1 , V2 (V2 > V1) be the diameters of two concentric circles and Z be the length of a chord of a circle which is tangent to the other circle, prove that : V22=Z2+V12${V_{2}}^{2}=Z^{2}+{V_{1}}^{2}$.

Solution:-

Let XY be a chord of a circle which touches the other circle at Z. Then Δ$\Delta$OZY is a right triangle.

∴          By Pythagoras Theorem,

OZ2$OZ^{2}$ + ZY2$ZY^{2}$= OY2$OY^{2}$

i.e.,        (V12)2$\left ( \frac{V1}{2} \right )^{2}$+ (Z2)2$\left ( \frac{Z}{2} \right )^{2}$= (V22)2$\left ( \frac{V2}{2} \right )^{2}$

$\Rightarrow$           V124$\frac{{V_{1}}^{2}}{4}$+ Z24$\frac{Z^{2}}{4}$=            V224$\frac{{V_{2}}^{2}}{4}$

$\Rightarrow$         V22+Z2=V12${V_{2}}^{2}+Z^{2}={V_{1}}^{2}$

2. Two tangents AB and AC are drawn from an external point to a circle with centre 0. Prove that BOCA is a cyclic quadrilateral.

Sol. We know that,  tangents to a circle is perpendicular to its radius at the point of contact.

∴     OC$\perp$AC and OB$\perp$AB

$\angle$OCA=$\angle$OBA=90$90^{\circ}$

$\angle$OCA+$\angle$OBA=90$90^{\circ}$+ 90$90^{\circ}$

=180$180^{\circ}$

Sum of opposite angles of quadrilateral BOCA is 180$180^{\circ}$

Hence, BOCA is a cyclic quadrilateral.