## Tangents are drawn to the circle $x^2 + y^2 = 50$ from a point ‘P’ lying on the x-axis…

Problem: Tangents are drawn to the circle $x^2 + y^2 = 50$ from a point ‘P’ lying on the x-axis. These tangents meet the y-axis at points ‘P1’ and ‘P2’. Possible coordinates of ‘P’ so that area of triangle PP1P2 is minimum, is / are

(A) (10, 0)

(B) (10√2 , 0)

(C) (-10, 0)

(D) (-10√2 , 0)

Sol.

OP = 5√2 secθ ,

OP1 = 5√2 cosecθ

Δ PP1P2 = 100/sin2θ

(Δ PP1P2)min = 100

⇒ θ = π/4 ⇒ OP =10

⇒ P=(10, 0), (–10, 0)

Hence (A) and (C) are the correct answers.

## The tangents drawn from the origin $x^2 + y^2 + 2gx + 2fy + f^2 =0$ are perpendicular If

Problem: The tangents drawn from the origin $x^2 + y^2 + 2gx + 2fy + f^2 =0$ are perpendicular If

(A) g = f

(B) g = -f

(C) g = 2f

(D) 2g = f

Sol. Since tangents drawn from origin are perpendicular that means origin lies on director circle of given circle.

$\sqrt{g^2 + f^2 } = \sqrt{2}\sqrt{g^2 + f^2 -f^2}$

$\large g^2 + f^2 = 2 g^2$

f = ± g.

Hence (A) and (B) are the correct answer.

## If the point (k + 1, k) lies inside the region bound by the curve $x = \sqrt{25-y^2}$ and the y-axis, then the integral value of k

Problem: If the point (k + 1, k) lies inside the region bound by the curve $x = \sqrt{25-y^2}$ and the y-axis, then the integral value of k is / are

(A) 0

(B) 1

(C) 2

(D) 3

Solution:

Since (k + 1, k) lies inside the region bounded by $x = \sqrt{25-y^2}$

⇒ (k + 1)2 + k2 – 25 < 0 and k + 1 > 0

⇒ 2k2 + 2k – 24 < 0 and k > –1

⇒ –4 < k < 3 and k > –1

⇒ –1 < k < 3

Hence (A), (B) and (C) are the correct answer.

## The common chord of $x^2 + y^2-4x-4y = 0$ and $x^2 + y^2 = 16$ subtends at the origin an angle equal to

Problem: The common chord of $x^2 + y^2-4x-4y = 0$ and $x^2 + y^2 = 16$ subtends at the origin an angle equal to

(A) π/6

(B) π/4

(C) π/3

(D) π/2

Sol. The equation of the common chord of the circles $x^2 + y^2-4x-4y = 0$ and $x^2 + y^2 = 16$ is

x + y = 4 which meets the circle x2 + y2 = 16 at points A(4,0) and B(0,4).

Obviously OA ⊥ OB.

Hence the common chord AB makes a right angle at the centre of the circle

x2 + y2 = 16

Hence (D) is the correct answer.

## If two circles $(x – 1)^2 + (y – 3)^2 = r^2$  and $x^2 + y^2 – 8x + 2y + 8 = 0$  intersect in two distinct points then

Problem: If two circles $(x – 1)^2 + (y – 3)^2 = r^2$  and $x^2 + y^2 – 8x + 2y + 8 = 0$  intersect in two distinct points then

(A) 2 < r < 8

(B) r < 2

(C) r = 2

(D) r > 2

Sol. Let d be the distance between the centres of two circles of radii r1 and r2.

These circle intersect at two distinct points if
|r1-r2 | < d < r1+r2

Here, the radii of the two circles are r and 3 and distance between the centres is 5.

Thus, |r-3| < 5 < r+3

⇒ -2 < r < 8 and r > 2

⇒ 2 < r < 8

Hence (A) is the correct answer.

## If P(2, 8) is an interior point of a circle $x^2 + y^2 -2x + 4y -p = 0$ which neither touches nor intersects the axes…

Problem: If P(2, 8) is an interior point of a circle $x^2 + y^2 -2x + 4y -p = 0$ which neither touches nor intersects the axes, then set for p is

(A) p < -1

(B) p < -4

(C) p > 96

(D) φ

Sol. For internal point p(2, 8), 4 + 64 – 4 + 32 – p < 0

⇒ p > 96

and x intercept $\large =2\sqrt{1+p}$  therefore 1 + p < 0

⇒ p < -1 and y intercept $\large =2\sqrt{4+p}$

⇒ p < -4

Hence (D) is the correct answer.

## Equation of chord AB of circle $x^2 + y^2 = 2$  passing through P(2 , 2) such that PB/PA = 3…

Problem: Equation of chord AB of circle $x^2 + y^2 = 2$  passing through P(2 , 2) such that PB/PA = 3, is given by

(A) x = 3 y

(B) x = y

(C) y – 2 = √3(x – 2)

(D) none of these

Sol. Any line passing through (2, 2) will be of the form $\large \frac{y-2}{sin\theta} = \frac{x-2}{cos\theta} = r$

When this line cuts the circle x2 + y2 =2 ,

(rcosθ + 2)2 + (r sinθ + 2)2 = 2

⇒  r2 + 4(sinθ+ cosθ)r + 6 = 0

$\large \frac{PB}{PA} = \frac{r_2}{r_1}$ , now if r1 = α , r2 = 3α ,

then 4α = – 4(sinθ + cosθ), 3α2 = 6

⇒ sin2θ = 1

⇒ θ = π/4

So required chord will be y – 2 = 1 ( x –2)

y = x.