Circles
Family of Circles
Grade 11
Question:
<p>The circle passing through the intersection of the circles, $x^2 + y^2 - 6x = 0$ and $x^2 + y^2 - 4y = 0$, having its centre on the line, $2x - 3y + 12 = 0$, also passes through the point</p>
<p>(a) $(-1, 3)$</p>
<p>(b) $(-3, 1)$</p>
<p>(c) $(1, -3)$</p>
<p>(d) $(-3, 6)$</p>
Step-by-Step Solution
Key Concept: Use the family of circles formula: circles passing through intersection of two circles can be expressed as $S_1 + \lambda S_2 = 0$. Find $\lambda$ using the constraint that the centre lies on the given line.
<p><strong>Solution:</strong></p><p>Equation of circle passing through the intersection of the circles $x^2 + y^2 - 6x = 0$ and $x^2 + y^2 - 4y = 0$ is:</p><p>$(x^2 + y^2 - 6x) + \lambda(x^2 + y^2 - 4y) = 0$ where $\lambda \neq -1$</p><p>$(1 + \lambda)x^2 + (1 + \lambda)y^2 - 6x - 4\lambda y = 0$</p><p>Centre is $C\left(\frac{3}{1+\lambda}, \frac{2\lambda}{1+\lambda}\right)$ and it lies on $2x - 3y + 12 = 0$</p><p>$2 \cdot \frac{3}{1+\lambda} - 3 \cdot \frac{2\lambda}{1+\lambda} + 12 = 0$</p><p>$\frac{6 - 6\lambda}{1+\lambda} + 12 = 0$</p><p>$6 - 6\lambda + 12(1 + \lambda) = 0$</p><p>$6 - 6\lambda + 12 + 12\lambda = 0$</p><p>$18 + 6\lambda = 0$</p><p>$\lambda = -3$</p><p>Equation of required circle: $2x^2 + 2y^2 + 6x - 12y = 0$</p><p>$x^2 + y^2 + 3x - 6y = 0$</p><p>Checking option (d): $(-3)^2 + 6^2 + 3(-3) - 6(6) = 9 + 36 - 9 - 36 = 0$ ✓</p><p>∴ Answer is (d) $(-3, 6)$</p>
Correct Answer: D