The Circle
ICSE · Class 11 · Mathematics
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A circle has centre (3, -2) and passes through the point (6, 2). What is its equation in standard form?
A circle is concentric with x^2 + y^2 - 6x + 12y + 15 = 0 and has twice its area. What is its equation in standard form?
The circles x^2 + y^2 - 4x + 6y + 8 = 0 and x^2 + y^2 - 10x - 6y + 14 = 0 touch externally. What is their point of contact?
The circles x^2 + y^2 + 2x - 6y + 9 = 0 and x^2 + y^2 + 8x - 6y + 9 = 0 touch internally. What is their point of contact?
Sample Questions
A circle has centre (0, 0) and passes through the points (3, 4) and (-4, 3). What is its equation?
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x^2 + y^2 = 25
Since the centre is (0, 0), the radius is the distance to (3, 4): sqrt(3^2 + 4^2) = 5. The same result is obtained from (-4, 3). Therefore the equation is x^2 + y^2 = 25.
A point P(x, y) is given by x = 2 + 3 cos θ and y = 3 sin θ, where 0 ≤ θ < 2π. What Cartesian equation does P satisfy?
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(x - 2)^2 + y^2 = 9
From x = 2 + 3 cos θ, we get (x - 2)/3 = cos θ. From y = 3 sin θ, we get y/3 = sin θ. Squaring and adding gives (x - 2)^2/9 + y^2/9 = 1, so (x - 2)^2 + y^2 = 9.
A point P(x, y) is given by x = -2 + 5 cos θ and y = 4 + 3 sin θ, where 0 ≤ θ < 2π. Which equation represents the locus of P?
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9(x + 2)^2 + 25(y - 4)^2 = 225
Rewrite as (x + 2)/5 = cos θ and (y - 4)/3 = sin θ. Squaring and adding gives (x + 2)^2/25 + (y - 4)^2/9 = 1, or 9(x + 2)^2 + 25(y - 4)^2 = 225. Since the coefficients of x^2 and y^2 are different, it is not a circle.
A circle has diameter endpoints (1, 3) and (4, 5). What is its equation?
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(x - 1)(x - 4) + (y - 3)(y - 5) = 0
Using the diameter equation with endpoints (x1, y1) = (1, 3) and (x2, y2) = (4, 5), the equation is (x - x1)(x - x2) + (y - y1)(y - y2) = 0. Substituting gives (x - 1)(x - 4) + (y - 3)(y - 5) = 0.
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