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# Geometry

CIRCLES PART-2

__CIRCLES PART-2__

__Tangents And Cyclic Quadrilateral__

__Tangent to a Circle:__

__Tangents from an External Point to a Circle:__

Thus, we can say that:

1. AP = BP i.e. lengths of tangents drawn from an external point are equal.

2. Angle APO = Angle BPO i.e. the line joining the external point and the centre of the circle acts as an angle bisector of the angle made by the two tangents.

3. Angle AOP = Angle BOP

Also if we join the two points A and B, then the chord AB is known as the chord of contact.

**Example 1****:** PQRS is a square. SR is a tangent (at point S) to the circle with centre O and TR = OS. Then the ratio of area of the circle to the area of the square is **( CAT 1996)**

**Solution: **As SR is the tangent at point S, so SR is perpendicular to OS. Also given that TR = OS. Also OS = OT

Consider ?OSR,

SR^{2} = OR^{2} – OS^{2} ( Pythagoras Theorem)

SR^{2} = 4r^{2} –r^{2 }= 3r^{2}

Or SR = √3r

**Area of circle/Area of square = **** = **

** Example2 : **ABCD is a rectangle and a circle is drawn tangent to the sides AD,CD and AC. If AB = 24 and AD = 10 units, then the length of OC is

**Solution: **

Here AC is the diagonal of the rectangle ABCD. So AC = 26 units

Consider ?ADC which is a right angled triangle and the given circle is the incircle of the above triangle.

In-radius of a right triangle = (a+b-c)/2 where a,b are the perpendicular sides while c is the length of the hypotenuse.

r = (10+24 -26)/2 = 4 units.

MC = 24-r = 24-4 = 20 units.

Consider ?OCM

OC^{2} = OM^{2} + MC^{2}

OC^{2 }= 4^{2} + 20^{2}

= 16 + 400 = 416

**OC = 4√26 units **

__Alternate Segment Theorem:__

**Solution**

**Tangents Common To Two Circles:**

__Case1 : Non Intersecting Circles:__

Case2: Two circles touch each other externally

Case3: Two intersecting circles:

Case4: Circles touch each other internally:

Case5: Circles are concentric

__Length of Direct Tangent and Transverse Tangent:__

**Example 4:** Two circles of radii 4 cm and 9 cm respectively touch each other externally at a point and a common tangent touches them at the points P and Q respectively. Then, the area of a square with side PQ, is

1. 97 sq. cm 2. 194 sq. cm 3. 72 sq. cm. 4. 144 sq. cm.

**Solution: **

**Example 5**. ABCD is a cyclic quadrilateral. Sides AB and DC, when produced meet at the point P and sides AD and BC, when produced meet at the point Q. If angle ADC = 65^{0} and angle BPC = 10^{0}, then angle CQD is equal to

1. 30^{0} 2. 40^{0} 3. 55^{0} 4. 85^{0}

Solution:

**Example 6:** In the given figure, AB is diameter of the circle and points C and D are on the circumference such that ∠CAD = 30° and ∠CBA = 70°. What is the measure of ∠ACD? **(CAT 1995)**

**Solution:** As AB is the diameter, so angle ACB = 90^{0}

As ABCD is a cyclic quadrilateral, so ∠B + ∠D = 1800

Or ∠ADC = 110^{0 }

Consider ?ADC,

∠CAD + ∠ADC + ∠ACD = 1800 ( Angle sum property)

So, ∠ ACD = 40^{0}.

**Example 7:** ABCD is a cyclic trapezium in which AB = 5 and BC = 3 units. If AB is the diameter of the circle, then what is the length of CD?

**Solution:** As ABCD is a cyclic trapezium, so it must be an isosceles trapezium as well i.e. the non parallel sides are equal in length.

So, AB = 5, BC = 3 and AD = 3units.

Join AC and BD. As AB is the diameter, so angle ACB = 900

Applying Pythagoras theorem in ?ACB we get

AC = BD = 4 units.

In cyclic quadrilateral,

AB*CD + BC*AD = BD*AC ( Ptolemy’s Theorem)

5x + 3*3 = 4*4

5x = 7 or x = 7/5

**Example 8:** In the figure below(not drawn to scale), rectangle ABCD is inscribed in the circle with centre at O. The length of side AB is greater than that of side BC. The ratio of the area of the circle to the area of the rectangle ABCD is π :√3. The line segment DE intersects AB at E such that ∠ODC = ∠ADE. What is the ratio of AE : AD?

- 1: √3 B. 1 : √2 C. 1 : 2√3 D. 1:2

**Solution: **

tanθ = AE/AD = (b/2)/(l/2) = b/l

Also r^{2} = (b/2)^{2} + (l/2)^{2 }

Area of circle/Area of rectangle = π :√3

πr^{2} /lb = π/3

Or r^{2} = lb/√3

b^{2}/4 + l^{2}/4 = lb/√3

√3b^{2} + √3l^{2} = 4lb

Divide by l^{2} on both sides

√3(b/l)^{2 }+ √3 = 4(b/l)

Let b/l = x

√3x^{2} -4x + √3 = 0

x = √3 or 1/√3

So, b/l = AE/AD = 1:√3

(Here x = √3 gets ruled out because x = b/l = tanθ giving θ = 60^{0}, which is not possible)