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.