# Circle

Co-ordinate Geometry

## If from any point on the circle x^2 + y^2 = a^2 tangents are drawn to the circle x^2 + y^2 = a^2 sin^2α , then the angle between the tangents, is

Q: If from any point on the circle x2 + y2 = a2 tangents are drawn to the circle x2 + y2 = a2 sin2 α , then the angle between the tangents, is

(A) α/2

(B) α

(C) 2α

(D) 4α

Sol:

The two given circles are concentric.

From figure, OP = a, OA = a sinα

If 2θ is the angle between the tangents,

$sin\theta = \frac{a sin\alpha}{a}$

⇒ θ = α

Required angle between PA and PB = 2 α

Hence (C) is the correct answer.

## The centre of the circle inscribed in a square formed by the lines x^2 – 8x + 12 = 0 and y^2 – 14y + 45 = 0 is

Q: The centre of the circle inscribed in a square formed by the lines x2 – 8x + 12 = 0 and y2 – 14y + 45 = 0 is

(A) (4 , 7)

(B) (7, 4)

(C) (9 , 4)

(D) (4 , 9)

Sol:

Centre is the midpoint of AC ≡ (4 , 7).

Hence (A) is the correct answer.

## The locus of a point from which the length of tangents to the circles x^2 + y^2 = 4 and 2(x^2 + y^2) – 10x + 3y – 2 = 0 are equal, is

Q: The locus of a point from which the length of tangents to the circles x2 + y2 = 4 and 2(x2 + y2) – 10x + 3y – 2 = 0 are equal, is

(A) a straight line inclined at π/4 with the line joining the centres of the circles

(B) a circle

(C) an ellipse

(D) a straight line perpendicular to the line joining the centres of the circles

Ans: (D)

## The co-ordinates of the point on the circle x^2 + y^2 – 12x – 4y + 30 = 0 which is farthest from the origin are

Q: The co-ordinates of the point on the circle x2 + y2 – 12x – 4y + 30 = 0 which is farthest from the origin are

(A) (9 , 3)

(B) (8, 5)

(C) (12 , 4)

(D) none of these

Sol: Centre of the given circle (6 , 2 )

equation of straight line passing through (0.0) & (6,2) will be

y = 3 x

Now put y= 3x in the equation of circle & solve

Ans : (A)

## Equation of a circle with centre (4, 3) touching the circle x^2 + y^2 = 1 is

Q: Equation of a circle with centre (4 , 3) touching the circle x2 + y2 = 1 is

(A) x2 + y2 – 8x – 6y – 9 = 0

(B) x2 + y2 – 8x – 6y + 11 = 0

(C) x2 + y2 – 8x – 6y – 11 = 0

(D) x2 + y2 – 8x – 6y + 9 = 0

Sol: Let the circle be (x – 4)2  + (y – 3)2 = r2

The point (4 , 3) lie outside x2 + y2 = 1  so they touch each other externally.

Hence , Sum of their radii = Distance between their centres

$r + 1 = \sqrt{(4-0)^2 + (3-0)^2}$

⇒ r = 4

∴  circle  is (x – 4)2 + (y – 3)2 = 42

⇒ x2 + y2 – 8x – 6y + 9 = 0

Hence (D) is the correct answer.