## 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.

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≡ $\displaystyle 2+ \frac{4}{\sqrt{13}} , 3 + \frac{6}{\sqrt{13}}$

## Find the equation of the circle of minimum radius which contains the three circles …..

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

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

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

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 (acosθi, a sinθi); 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 .

## The centre of the circle S = 0 lies on the line 2x – 2y + 9 = 0 and S = 0 cuts orthogonally the 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).

## Find the locus of the foot of the perpendicular let fall from the origin upon any chord of the circle …

Q: Find the locus of the foot of the perpendicular let fall from the origin upon any chord of the circle x2 + y2 + 2gx + 2fy + c = 0 which subtends a right angle at the origin.

Sol. Given circle is x2 + y2 + 2gx + 2fy + c = 0 … (i)

let the equation of any chord be lx + my = 1 … (ii)

making (i) homogeneous with the help of (ii) we get,

x2 + y2 + (2gx + 2fy) (lx + my) + c (lx + my)2 = 0

It represents two straight lines passing through origin

⇒ 1 + 2gl + cl2 + 1 + 2fm + cm2 = 0

⇒ 2 + 2(gl + fm) + c (l2 + m2) = 0 … (iii)

Let the perpendicular from the origin on (ii) meet it at P, whose coordinates are

(h, k), equation of this perpendicular line will be ly – mx = 0 … (iv)

as (h, k) is the point of intersection, it will satisfy (ii) and (iv)

Hence, lh + mk – 1 = 0 … (v)

and lk – mh = 0 … (vi)

Eliminating l and m we get,

$\displaystyle l = \frac{h}{h^2 + k^2}$ and $\displaystyle m = \frac{k}{h^2 + k^2}$ putting these values in (iii) we get,

required locus as 2x2 + 2y2 + 2gx + 2fy + c = 0; which is a circle