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Solved Subjective Questions on Circle Set 3

Posted on - 11-02-2017

JEE Math Circle

IIT JEE

Example 1.

A rectangle ABCD is inscribed in a circle with a diameter lying along the line
3y = x + 10
. If A and B are the points (-6, 7) and (4, 7) respectively, find the area of the rectangle.

Solution

Let O (h, k) be the centre of circle.
OA2 = OB2.


&⇒
(h + 6)2 + (k – 7)2 =(h–4)2 +(k – 7)2


&⇒
h2 +36 + 12h = h2 + 16 – 8h


&⇒
20h = –20
&⇒
h = –1

(h, k) also lies on the line 3y = x + 10


&⇒
3k = h + 10
&⇒
3k = 9


&⇒
k = 3
&⇒
O º (-1, 3)

Now AB =,

OB = = and BD = .

In DADB, BD2 = AD2 + AB2 i.e. 164 = AD2 + 100.


&⇒
AD2 = 64
&⇒
AD = 8

Area of Rectangle ABCD = 10 ´ 8 = 80 sq. units.

Example 2.

AB is the diameter of a circle, CD is a chord parallel to AB and 2 CD = AB. The tangent at B meets the line AC produced at E. Prove that AE = 2AB.

Solution

Let O, the centre of circle, be taken as the

origin. A (-a, 0) and B (a, 0) a is radius of the .

circle. In DOCF,.

OC2=CF2+OF2 (Since CD=AB and CD|| AB)


&⇒
a2 = + OF2
&⇒
OF2 =


&⇒
OF = ±
&⇒
C

Equation of the line AE where A (-a, 0), is y – 0 =

y = Ö3 (x + a) … (1)

Equation of line BE [which is tangent at (a, 0) is

x = a … (2)

Point of intersection of (1) and (2) is E i.e. E (a, 2Ö3a).

AE = = 4a = 2(2a) = 2AB

Hence AE = 2AB

Example 3.

Find the locus of centres of the circle which touches the two circle x2 + y2 = a2 and x2 + y2 = 4ax externally. .

Solution

Let (h, k) be the centre of the circle and r be its radius then = r + a

and

Eliminating r,

\ locus of (h, k) is

Squaring (x – 2a)2 + k2 = a2 + x2 + y2 + 2a


&⇒
- 4ax + 3a2 = 2a

Squaring we get 12 x2 – 4y2 – 24ax + 9a2 = 0.

Example 4.

If i = 1, 2, 3, 4 mi > 0 are 4 distinct points on a circle, then show that
m1 .m2 .m3 .m4 = 1.

Solution

where i = 1, 2, 3, 4 are four points lying on a circle.

Let the equation of the circle be x2 + y2 + 2gx + 2fy + c = 0

is lying on the circle


&⇒
mi2 + + 2gmi + = 0


&⇒
mi4 + 2gmi3 + 2fmi + cmi2 + 1 = 0

If m1, m2, m3, m4 are its roots then
&⇒
m1 × m2 × m3 × m4 = 1

Example 5.

If al2 – bm2 + 2dl + 1 = 0 a, b, d are fixed real numbers such that a + b = d2, then prove that the line lx + my + 1 = 0 touches a fixed circle. Find its equation.

Solution

Let the equation of the circle be x2 + y2 + 2gx + 2fy + c = 0

If lx + my + 1 = 0 is a tangent to the circle

Then


&⇒
(g2 + f2 – c) (l2 + m2) = (lg + mf – 1)2


&⇒
(f2 – c) l2 + (g2 – c)m2 + 2l (g – mgf) + 2mf – 1 = 0

Comparing with the given condition al2 – bm2 + 2dl + 1 = 0,

– (f2 – c) = a, – (g2 – c) = – b, -g (1 – mf) = d, f = 0


&⇒
c = a, g = – d, g2 – c = b, d2 – a = b


&⇒
a + b = d2 which is the required condition.

Hence the fixed circle is x2+ y2 – 2dx + a = 0.

 
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