**1.** From a point Q, the length of the tangent to a circle is 24 cm and distance of Q from the center is 25 cm. The radius of the circle is

a) 7 cm

b) 12 cm

c) 15 cm

d) 24.5 cm

**Answer:** 7

**Solution: **

Since P is the point of contact

[latex]\Rightarrow[/latex] [latex]\angle[/latex]OPQ = 90

[latex]\Rightarrow[/latex] [latex]OP^{2}+PQ^{2}=OQ^{2}[/latex]

[latex]\Rightarrow[/latex] [latex]OP^{2}+24^{2}=25^{2}[/latex]

[latex]\Rightarrow[/latex] [latex]OP^{2}=49[/latex]

[latex]\Rightarrow[/latex] OP = 7

**2. **Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

**Proof:** Let AB is a diameter and PR and QS are the tangents at point A and B respectively.

Since A and B are the Point of contacts of Tangent PR and QS respectively

[latex]\Rightarrow[/latex] [latex]\angle[/latex]PAB = 90 and [latex]\angle[/latex]QBA = 90

[latex]\Rightarrow[/latex] [latex]\angle[/latex]PAB + [latex]\angle[/latex]QBA = 90 + 90 = 180

Since sum of interior angles between two line is 180

hence, PR is parallel to QS

**3. **A quadrilateral ABCD is drawn to circumscribe a circle as shown in the figure. Prove that AB + CD = AD + BC

**Proof:** AP = AS (1)

PB = BQ (2)

CR = RQ (3)

DR = RS (4)

{ tangents drawn from external point are equal }

Add Equation (1) , (2) , (3) and (4) , we get

AP + PB + CR + DR = AS + BQ + RQ + RS

AB + CD = AD + BC

**4.** A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively as shown in figure. Find the sides AB and AC.

Solution: Construction: Draw line OE, OF, OA, OB, OC

CD = CE = 6

BD = BF = 8

AE = AF = x

s = (6 + 6 + 8 + 8 + x + x) / 2 = 14 + x

Area = [latex]\sqrt{(14+x)(x)\times8\times6}[/latex] = [latex]\frac{1}{2}\times4\times(28+2x)[/latex]

Squaring on both sides, we get

[latex](14+x)(x)\times8\times6[/latex] = [latex]4\times4\times(14+x)^{2}[/latex]

48x = 16(14 + 2x)

x = 14

AB = 14 + 8 = 22

AC = 14 + 6 = 20

**5. **Prove that the parallelogram circumscribing a circle is a rhombus.

**Proof: **

Since ABCD is a quadrilateral (parallelogram) , therefore AB + CD = AD + BC

2AB = 2AD ( AB = CD and AD = BC )

AB = AD

Since adjacent sides of a parallelogram are equal

hence, it is a rhombus.