Solved Subjective Questions on Circle Set 1

Example 1.

Find the equation of the circle with centre on the line 2x + y = 0 and touching the lines 4x - 3y + 10 = 0 and 4x - 3y - 30 = 0.

Solution

Since the two lines are tangents to the given circle and they are parallel, we have

Note: We know that if ax + by + c = 0 and ax + by + c¢ = 0are two parallel tangents to a circle, then equation of the line parallel to given lines and passing through the centre is given by ax + by + .

Hence the centre of the circle lies on 4x - 3y - 10 = 0

Also the centre lies on 2x + y = 0

Hence the coordinates of the centre are (1, -2).

&⇒ The equation of the circle is (x - 1)² + (y + 2)² = 16.

Example 2

Find the locus of the mid-points of the chords of the circle
x² + y² - 2x - 6y - 10 = 0 which pass through the origin.

Solution

Let (h, k) be the coordinates of the mid point.

Equation of the chord whose mid point is (h, k) is

xh + yk - (x + h) - 3(y + k) - 10 = h² + k² - 2h - 6k - 10 (using T = S₁)

i.e., h² + k² - h(x + 2) - k(6 + y) + x + h + 3y + 3k = 0.

This chord passes through the origin (0, 0). .

Hence h² + k² - h - 3k = 0

Hence the required locus is x² + y² - x - 3y = 0.

Alternative:

If m₁ is the slope of the chord, then

If m₂ is the slope of the perpendicular to the chord then where (1, 3) is

the centre of the given circle. But m₁m₂ = -1 .

&⇒ h² + k² - h - 3k = 0

Hence the required locus is x² + y² – x – 3y = 0.

Example 3.

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+12xy -16y-24x=9x²+ 9y²–36.

&⇒ 5y² + 24x + 16y - 12xy - 52 = 0

Here a = 0, b = 5 h = -6. .

Hence angle q between the tangents is given by

or tanq =

or tanq =
&⇒ q = tan^-1.

Hence the angle between the pair of tangents is

tan^-1.

Example 4.

Find the locus of middle points of chords of the circle (x–2)² + (y –3)² = a², which subtends a right angle at the point (b, 0).

Solution

Let M (h, k) be the middle points of the chord AB. Also ∠APB = 90° and AM = PM.

We have AM =

PM =

&⇒ b²– 2bh+h² + k² = a² – (h–2)² – (k – 3)²

&⇒ h²+k²–h(b–2)– 3k +

Hence locus of (h, k) is

x² + y² – x (b – 2) – 3y +

which is a circle.

Example 5.

If a straight line through C(-Ö8, Ö8), making an angle of 135° with the x-axis, cuts the circle x = 5 cos q, y = 5 sin q, in points A and B, find the length of the segment AB.