## The equation of the circle passing through (0, 1) and touching the curve y = x^2 at (2, 4) is …..

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.

## If tangents be drawn to the circle x^2 + y^2 = 12 at its points of intersection, with the circle x^2 + y^2 – 5x + 3y – 2 = 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)

## 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 ….

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.

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

Q: If (mi , 1/mi) i = 1, 2, 3, 4 mi > 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.

## 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 ….

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

## Find the equations of the tangents from the point A(3, 2) to the circle x^2 + y^2 = 4 and hence find the angle between …

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

## If the chord of contact of the circle x^2 + y^2 = b^2 generated by a point on the circle x^2 + y^2 = a^2 touches the circle ….

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.

## 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 ….

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.

## Two rods of length a and b slide along the axes, which are rectangular, in such a manner that their ends are concyclic….

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

## The abscissae of two points A and B are the roots of the equation x^2 + 2ax – b^2 = 0 and their ordinates are the roots of ….

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

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