QuizSolver
  • Bank PO
  • CBSE
  • IIT JEE
 
 

Solved Subjective Questions on Circle Set 7

Posted on - 21-02-2017

JEE Math Circle

IIT JEE

Example 1.

Let y = mx be the equation of a chord of a circle whose radius is ‘a’, the origin of coordinates being one extremity of the chord, and the axis of x being a diameter of the circle. Prove that the equation of the circle of which this chord is a diameter is (1 + m2) (x2 +y2) –2a(x+ my) = 0. .

Solution

(a, 0) is centre of given circle whose equation is (x – a)2 + y2 = a2

i.e. x2 + y2 – 2ax = 0 .

Equation of circle with OP (y = mx) as diameter is

x2 + y2 – 2ax + l(y – mx) = 0

whose centre is

This centre must lie on y = mx

\
&⇒
l = -

\ centre is x2 + y2 – 2ax - (y – mx) = 0

i.e. (1 + m2) (x2 + y2) – 2a(1 + m2)x – 2amy + 2am2x = 0 .

i.e. (1 + m2)(x2 + y2) – 2ax – 2amy = 0.

Example 2.

C is a circle with centre (0, d) and radius r (r < d). A point P is chosen on the x-axis and a circle is drawn with centre at P which touches C externally and meets the x-axis in points M and N. Find the coordinates of a point Q on the y-axis such that ∠MQN is constant for any choice of the point P.

Solution

Suppose P º (h, 0) and Q º (0, k) and the radius is s.


&⇒
M º (h - s, 0) and N º (h + s , 0)
. .

Then (r + s)2 = d2 + h2

Also, slope of the line

Slope of line NQ = .
Also q = q2 - q1

Put (r + s)2 = d2 + h2 to eliminate h


&⇒

If we take k2 = d2 - r2, then

Hence for k2 = d2 - r2, ∠MQN is constant for all points P.

Example 3.

Show that the equation to the circle cutting orthogonally the circles
(x –a)2 + (y – b)2 = b2, (x – b)2 + (y – a)2 = a2 and (x – a –b – c)2 + y2 = ab + c2 is
x2 + y2 – 2x (a + b) – y (a + b) + a2 + 3ab + b2 = 0.

Solution

The three circles are x2 + y2 – 2ax – 2by + a2 = 0

x2 + y2 – 2bx – 2ay + b2 = 0

x2 + y2 – 2(a+ b + c)x + a2 + b2 + ab + 2bc + 2ac = 0

Let x2 + y2 + 2gx + 2fy + k = 0

Cut them orthogonally, then

2g(-a) + 2f(-b) = k + a2 ….(1).

2g(-b) + 2f(-a) = k + b2 ….(2).

2g(-a – b - c) + 2f.0 = k + a2 + b2 + ab + 2c(a + b) ….(3).

Eliminating f from (1) and (2), we get

- 2g(a + b) = k + a2 + ab + b2

Substituting from (3), - 2gc = 2c(a + b)
&⇒
- (a + b) = g

\ k = 2(a + b)2 – a2 – ab – b2 = 2(a2 + 2ab + b2) – a2 – ab – b2

= a2 + b2 + 3ab

\ (1)
&⇒
f = -

so the equation of required circle is

x2 + y2 – 2(a + b)x – (a + b)y + a2 + b2 + 3ab = 0

Example 4.

Is (3, 2) an interior point or an exterior point of the circle x2 + y2 – 2x + y = 0. If interior, find the equation of the circle centred on it and of maximum area contained in the given circle. If exterior find the equation of the circle of minimum radius centred on it containing the given circle. .

Solution

32 + 22 – 6 + 4 > 0 so (3, 2) is exterior point of the circle

x2 + y2 – 2x + y = 0 …(1)

The radius of required circle

= distance between centres (3, 2) and (1, -1/2) + radius of circle (1)

= =

\ Equation of required circle is (x – 3)2 + ( y – 2)2 =

Example 5.

If ABCD is a cyclic quadrilateral, prove that the orthocentre of the triangle ABC, BCD, CDA and DAB lies on a circle.

Solution

Let (acosqI, a sinqi); i = 1, 2, 3, 4 represents the points A, B, C and D respectively. .

The orthocentre of DABC is (a(cosq1 + cosq2 + cosq3), a (sinq1 + sinq2 + sinq3))

This will lies on the circle

[x - a(cosq1 + cosq2 + cosq3)]2 + [y – a(sinq1 + sinq2 + sinq3)]2 = a2

because (a cosq4)2 + (a sinq4)2 = a2 is true

Similarly the orthocentre of DBCD, DACD, DABD all lie on circle (1).

 
Quadratic Equations - Solved Objective Questions Part 2 for Conceptual Clarity
Quadratic Equations - Solved Objective Questions Part 1 for Conceptual Clarity
Solved Objective Question on Probability Set 2
Solved Objective Question on Probability Set 1
Solved Objective Question on Progression and Series Set 2
Solved Objective Question on Permutations and Combinations Set 3
Solved Objective Question on Permutations and Combinations Set 2
Solved Objective Question on Progression and Series Set 1
Solved Objective Question on Permutations and Combinations Set 1
Quadratic Equations - Solved Subjective Questions Part 4
Quadratic Equations - Solved Subjective Questions Part 2
Quadratic Equations - Solved Subjective Questions Part 3
Quadratic Equations - Solved Subjective Questions Part 1
Solved Subjective Questions on Circle Set 9
Solving Equations Reducible to Quadratic Equations
Theory of Polynomial Equations and Remainder Theorem
Solved Subjective Questions on Circle Set 8
Solving Quadratic Inequalities Using Wavy Curve Methods
Division and Distribution of Objects - Permutation and Combination
Basics of Quadratic Inequality or Inequations

Comments