Circle

Q: The equation of the circle passing through (0, 1) and touching the curve y = x² at (2 , 4) is ……

Solution : The equation of tangent to the curve y = x² at (2, 4) is 4x –y = 4.

Family of circles touching the curve y = x² at (2, 4) is

(x – 2)² + ( y – 4)² + λ(4x – y –4) = 0.

Since the required circle passes through (0 , 1), we have

4 + 9 – 5 λ = 0 ⇒  λ = 13/5

Hence equation of the required circle is

(x –2)² + (y – 4)² + (13/5)(4 x – y -4) = 0

⇒ 5x² + 5y² + 32 x – 53 y + 48 = 0.

Q: If tangents be drawn to the circle x² + y² = 12 at its points of intersection, with the circle x² + y² – 5x + 3y – 2 = 0, then the coordinates of the point of intersection of the tangents is ……

Solution : Let the tangents to the circle x² + y² = 12 meet at the point (x1, y1)

Equation of the chord of contact is x x₁ + y y₁ – 12 = 0 ….(i)

Equation of the common chord of two circles

x² + y² – 12 = 0 and x² + y² – 5x + 3y – 2 = 0 is –5x + 3y +10 = 0 …. (ii)

since (i) and (ii) are identical,

$\displaystyle \frac{x_1}{-5} = \frac{y_1}{3} = \frac{-12}{10}$

⇒ x₁ = 6 and y₁ = -18/5

Hence the required point is (6 , -18/5)

Q: The centre 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 √2 units, then the equation of the circle in the new position is ………

Solution : $\displaystyle \frac{x-1}{cos\theta} = \frac{y-1}{sin\theta} = \pm r$

Since , y = x has slope tanθ = 1 , θ = π/4

$\displaystyle \frac{x-1}{1/\sqrt{2}} = \frac{y-1}{1/\sqrt{2}} = \pm \sqrt{2}$

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.

Q: If (m_i , 1/m_i) i = 1, 2, 3, 4 m_i > 0 are 4 distinct points on a circle, then show that m₁ .m₂ .m₃ .m₄ = 1.

Solution : $\displaystyle (m_i , \frac{1}{m_i}) $ ; where i = 1, 2, 3, 4 are four points lying on a circle.

Let the equation of the circle be x² + y² + 2 g x + 2 f y + c = 0

is lying on the circle $\displaystyle (m_i , \frac{1}{m_i}) $

$\displaystyle m_i^2 + \frac{1}{m_i^2} + 2 g m_i + 2 f \frac{1}{m_i} + c = 0 $

⇒ $\displaystyle m_i^4 + 2 g m_i^3 + 2 f m_i + c m_i^2 + 1 = 0 $

If m₁ , m₂, m₃, m₄ are its roots

then ⇒ m₁ × m₂ × m₃ × m₄ = 1.

Q: 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 = 4 b y

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

$\displaystyle C (\frac{x_1 + a}{2} , \frac{y_1 + b}{2}) $

and $\displaystyle Radius = \frac{y_1 + b}{2} $

$\displaystyle (\frac{x_1 + a}{2}-a)^2 + (\frac{y_1 + b}{2}-b)^2 = (\frac{y_1 + b}{2})^2$

$\displaystyle (\frac{x_1 – a}{2})^2 + (\frac{y_1 – b}{2})^2 = (\frac{y_1 + b}{2})^2$

$\displaystyle x_1^2 + a^2 – 2 a x_1 – 2 b y_1 = 2 b y_1 $

$\displaystyle x_1^2 + a^2 – 2 a x_1 – 4 b y_1 = 0 $

(x₁ –a)² = 4 b y₁

Hence the locus of (x₁, y₁) is

(x – a)² = 4 b y

Q: Find the equations of the tangents from the point A(3 , 2) to the circle x² + y² = 4 and hence find the angle between the pair of tangents.

Solution : The equation of the pair of tangents from (3, 2) is given by

T₂ = SS₁

⇒ (3x + 2y – 4)2 = (x² + y² – 4)(9 + 4 – 4)

i.e., 9x² + 4y² + 16 + 12 x y -16 y-24 x = 9x² + 9y² –36

⇒ 5y² + 24x + 16y – 12xy – 52 = 0

Here a = 0, b = 5 h = -6.

Hence angle θ between the tangents is given by

$\displaystyle tan\theta = \frac{2\sqrt{h^2 – a b}}{a+b}$

$\displaystyle tan\theta = \frac{2\sqrt{36}}{5} = \frac{2 \times 6}{5}$

$\displaystyle tan\theta = \frac{12 }{5}$

$\displaystyle \theta = tan^{-1} (\frac{12 }{5}) $

Hence the angle between the pair of tangents is $\displaystyle tan^{-1} (\frac{12 }{5}) $

Q: If the chord of contact of the circle x² + y² = b² generated by a point on the circle x² + y² = a² touches the circle x² + y² = c² , prove that a, b, c are in G.P.

Solution : The line $ y = m x + c \sqrt{1 + m^2}$…(i)

touches the circle x² + y² = c² for all m

Let (a cosθ , a sinθ) be a point on x² + y² = a² .

The equation of the chord of contact of the point with respect to the circle x² + y² = b² is

a x cosθ + a y sinθ = b² …. (ii)

Since (i) and (ii) represent the same line

$\displaystyle sin\theta = – \frac{acos\theta}{m} = \frac{b^2}{c\sqrt{1+m^2}}$

⇒ m = -cotθ and $ b^2 = ac \sqrt{1+m^2} sin \theta $

⇒ b² = ac

⇒ a , b , c are in G.P.

Q: Suppose f(x, y) = 0 is the equation of the circle such that f(x , 1) = 0 has equal roots (each equal to 2) and f(1, x) = 0 also has equal roots (each equal to 0). Find the equation of the circle.

Solution : Let f(x, y) = x² + y² + 2 g x + 2 f y + c

⇒  f(x, 1) = x² + 1 + 2 g x + 2f + c ≡ (x – 2)² (given)

⇒ g = -2 , 2 f + c + 1 = 4 . . .(i)

Also f(1 , x) = 1 + x² + 2g + 2fx + c ≡ x² (given)

⇒ f = 0 , 2 g + c + 1 = 0 . . .(ii)

solving, we get c = 3 ,  g = -2.

Thus equation of circle is x² + y² – 4x + 3 = 0.

Q: Two rods of length a and b slide along the axes, which are rectangular, in such a manner that their ends are concyclic. Prove that the locus of the centres of the circle passing through these ends is the curve 4(x² – y²) = a² – b²

Solution : AB = a , CD = b, AL = a/2 , CM = b/2

PL = k , PM = h

ln Δ PAL , PA² = PL² + AL²

$\displaystyle r^2 = k^2 + \frac{a^2}{4}$ …(i)

ln Δ PCM, PC² = PM² + CM²

$\displaystyle r^2 = h^2 + \frac{b^2}{4}$ …(ii)

From (i) and (ii), we get,

$\displaystyle k^2 + \frac{a^2}{4} = h^2 + \frac{b^2}{4}$

$\displaystyle \frac{a^2}{4} – \frac{b^2}{4} = h^2 – k^2 $

$\displaystyle a^2 – b^2 = 4 (h^2 – k^2) $

Locus of (h, k) is

4(x² – y²) = a² – b²

Q: The abscissae of two points A and B are the roots of the equation x² + 2 a x – b² = 0 and their ordinates are the roots of the equation x² + 2 p x – q² = 0. Find the equation and the radius of the circle with AB as diameter.

Solution : Let the given points A and B be A = (x₁ , y₁), B = (x₂, y₂)

x₁ + x₂ = -2a and x₁ x₂ = -b²

Also y₁ + y₂ = -2 p and y₁ y₂ = -q²

Equation of the circle is (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0

x² – (x₁ + x₂)x + x₁ x₂ + y² – (y₁ + y₂)y + y₁ y₂ = 0

x² + 2ax – b² + y² + 2py – q² = 0

x² + y² + 2ax + 2py – b² – q² = 0

$\displaystyle Radius = \sqrt{a^2 + p^2 + b^2 + q^2}$