Circle

Q: Find the point P on the circle x^2 + y^2 – 4x – 6y +9 = 0 such that (i) ∠POX is minimum (ii) OP is maximum, where O is origin and OX is the x-axis.

Sol. Given circle is

x² + y² – 4x – 6y + 9 = 0  …. (i)

Its centre is C(2,3) and radius is 2.

(i) Let OP and ON be the two tangents from O to circle (i), then ∠POX will be minimum when OP is tangent to the circle at P. Let ∠POX = θ , $\displaystyle OP = \sqrt{(x^2 + y^2 -4x -6y +9)_{0,0}}$

= 3 = ON

∠ COP = (90-θ)/2 ,

$\displaystyle cos(\frac{90-\theta}{2})= \frac{OP}{OC} = \frac{3}{\sqrt{13}}$

$\displaystyle sin(\frac{90-\theta}{2})= \frac{2}{\sqrt{13}}$

sin (90 – θ) = cosθ = 12/13 and sinθ = 5/13

P ≡ (OP cosθ , OP sinθ) i.e. P≡ (36/13 , 15/13)

(ii) OP will be maximum , if P becomes the point where extended part of OC cuts the circle. Let this point be Q. Then maximum value of OP = OQ = OC + radius $=\sqrt{13} + 2 $

slope of line OC $\displaystyle = \frac{3-0}{2-0} = \frac{3}{2} = tan\alpha $ (say)

$\displaystyle cos\alpha = \frac{2}{\sqrt{13}} $ , $\displaystyle sin\alpha = \frac{3}{\sqrt{13}} $

For maximum OP,

P ≡ (OQ cosα , OQ sinα )

i.e. P≡ $\displaystyle 2+ \frac{4}{\sqrt{13}} , 3 + \frac{6}{\sqrt{13}}$

Q: Find the equation of the circle of minimum radius which contains the three circles x2 + y2 – 4y – 5 = 0, x2 + y2 + 12x + 4y + 31 = 0 and x2 + y2 + 6x + 12y + 36 = 0.

Sol. For Ist circle centre (0, 2) and radius

$\displaystyle = \sqrt{(-2)^2 + 5} = 3 $

For IInd circle centre (–6, –2) and radius

$\displaystyle = \sqrt{6^2 + 2^2 -31} = 3 $

For IIIrd centre (–3, –6) and radius

$\displaystyle = \sqrt{3^2 + 6^2 -36} = 3 $

Let P(α , β) be the centre of the circle passing through the centres of the three

given circles. Then its radius r is given by

r² = α² + (β – 2)² = (α + 3)² + (β + 6)²

= (α + 6)² + (β + 2)²

or 4 – 4β = 6α + 9 + 12β + 36 = 12 α + 36 + 4β + 4

6α + 16 β + 41 = 0 …(i)

6α – 8β – 5 = 0 …(ii)

α = -31/18 , β = -23/12

The centre of the circle passing through centre of given circles is (-31/18 , -23/12)

and radius $\displaystyle = \sqrt{\alpha^2 + (\beta -2)^2}$

$\displaystyle = \sqrt{(-31/18)^2 + (\frac{-23}{12} -2)^2}$

$\displaystyle = \frac{5}{36}\sqrt{949} $

the radius of the required circle $\displaystyle = 3 + \frac{5}{36}\sqrt{949} $ and

centre (-31/18 , -23/12) . Equation of the required circle is

$\displaystyle (x+\frac{31}{8})^2 + (y+\frac{23}{12})^2 = (3 + \frac{5}{36}\sqrt{949})^2$
.

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

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

The orthocentre of DABC is

(a(cosθ₁ + cosθ₂ + cosθ₃), a (sinθ₁ + sinθ₂ + sinθ₃))

This will lies on the circle

[x – a(cosθ₁ + cosθ₂ + cosθ₃ + cosθ₄)]² + [y – a(sinθ₁ + sinθ₂ + sinθ₃ + sinθ₄)]² = a²

because (a cosθ₄)² + (a sinθ₄)² = a² is true

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

Q: The centre of the circle S = 0 lies on the line 2x – 2y + 9 = 0 and S = 0 cuts orthogonally the circle x² + y² = 4. Show that S = 0 passes through two fixed points and find their coordinates.

Solution: Let the circle S = 0 be represented by

x² + y² + 2gx + 2fy + c = 0

Its centre is (-g, -f)

⇒ 2(-g) -2(-f) + 9 = 0

⇒ 2g – 2f = 9 ….(i)

The circle S = 0 cuts the circle x2 + y2 – 4 = 0 orthogonally. Hence

2g × 0 + 2×f×0 = c – 4 Þ c = 4 …(2)

⇒ S ≡ x² + y² + (9 + 2f)x + 2fy + 4 = 0

⇒ x² + y² + 9x + 4 + 2f(x + y) = 0

This passes through the points of intersection of the line

x + y = 0 and the circle x² + y² + 9x + 4 = 0

⇒ x² + (-x)² + 9x + 4 = 0

⇒ 2x² + 9x + 4 = 0

⇒ x = -1/2, -4

⇒ The points are (–1/2, 1/2) and (–4, 4).