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Solved Subjective Questions on Circle Set 2

Posted on - 10-02-2017

JEE Math Circle

IIT JEE

Example 1.

Find the equation of the tangent at A to the circle ( x - a)2 + (y - b)2 = r2, where the radius through A makes an angle a with the x - axis.

Solution

The parametric equations of the circle

(x–a)2 + (y – b)2 = r2 are x = a + rcosa and y = b + rsina

Then and
&⇒

Now the equation of the tangent at (a + rcosa, b + rsina) to the given circle is
y – (b + rsina) = – cot a [x – (a + rcosa)]

or, (y – b) – rsina = – (x – a) cota + rcos a cot a

or, (x – a) cota + (y – b) = r [sina + cosa × cota]

or, (x – a) cosa + (y – b) sina = r

Alternate:

Equation of circle (x–a)2 + (y – b)2 = r2

Tangent at (x,y) = (a + rcosa, b + rsina) is

(x – a) (r cosa + a – a) + (y – b) (r sina + b – b) = r2


&⇒
(x – a) cosa + (y – b)sina = r

Example 2.

From the point A (0, 3) on the circle x2 + 4x + (y - 3)2 = 0 a chord AB is drawn and extended to a point M, such that AM = 2AB. Find the equation of locus of M.

Solution

Here AM = 2AB
&⇒
AB =


&⇒
B is the mid point of AM
. Let M be (h, k).

Hence B º

But AB is the chord of the circle x2 + 4x + (y – 3)2 = 0 and point B lies on the circle. .

Hence + 4 × + or,

or, or, h2 + 8h + k2 + 9 – 6k = 0

or, h2 + k2 + 8h – 6k + 9 = 0
&⇒
Locus of M is x2 + y2 + 8x – 6y + 9 = 0

Example 3.

Let A be the centre of the circle. x2 + y2 - 2x - 4y - 20 = 0. Suppose that the tangents at the points B (1, 7) and D (4, -2) on the circle meet at the point C. Find the area of the quadrilateral ABCD.

Solution

Equation of the circle is x2 + y2 – 2x – 4y – 20 = 0 where centre is A (1, 2) and

radius = AB = AD =

Equation of the tangent at B (1, 7) is

x + 7y – (x + 1) – 2(y + 7) – 20 = 0

Or x + 7y – x – 1 – 2y – 14 – 20 = 0


&⇒
5y = 35
&⇒
y = 7 … (1)

Equation of the tangent at D (4, –2) is

4x – 2y – (x + 4) – 2(y ­– 2) – 20 = 0

or 4x – 2y – x – 4 – 2y + 4 – 20 = 0

or 3x – 4y – 20 = 0 …(2)

Point of intersection of line (1) and (2) is C given by

3x – 28 – 20 = 0 or, 3x = 48 or, x = 16

C (16, 7), BC = =

DC = 15

Area of quadrilateral ABCD = Area D ABC+ Area of D ADC

= = 75. Sq.units.

Example 4.

The cente of a circle is (1, 1) and its radius is 5 units. If the centre is shifted along the line y – x = 0 through a distance Ö2 units. Find the equation of the circle in the new position. .

Solution

y = x has slope tanq = 1, q =

Hence

So centre in the new position are (2, 2), (0, 0). .

Equation of circle are x2 + y2 = 25 and (x – 2)2 + (y – 2)2 = 25

Example 5.

A variable circle passes through the point A (a, b) and touches the x - axis. Show that the locus of the other end of the diameter through A is (x - a)2 = 4by.

Solution

Let the other end of the diameter be(x1, y1). The centre and the radius of the required circle are given by Cand Radius =


&⇒


&⇒


&⇒


&⇒


&⇒
(x1–a)2=4by1
. .

Hence the locus of (x1, y1) is


&⇒
(x - a)2 = 4by

 
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