Circles
Family of Circles
Grade 11
Question:
<p>Consider the circle \(x^2 + y^2 - 10x - 6y + 30 = 0\). Let <i>O</i> be the centre of the circle and tangent at \(A(7, 3)\) and \(B(5, 1)\) meet at <i>C</i>. Let <i>S</i> = 0 represents family of circles passing through <i>A</i> and <i>B</i>, then:</p>
<p>(a) Area of quadrilateral OACB = 4</p>
<p>(b) the radical axis for the family of circles <i>S</i> = 0 is \(x + y = 10\)</p>
<p>(c) the smallest possible circle of the family <i>S</i> = 0 is \(x^2 + y^2 - 12x - 4y + 38 = 0\)</p>
<p>(d) the coordinates of point <i>C</i> are (7, 1)</p>
Step-by-Step Solution
Key Concept: The family of circles passing through two fixed points A and B has the form (given circle) + λ(radical axis) = 0. The smallest circle in this family has its diameter as the chord AB, and the radical axis is the common chord through A and B.
<p><strong>Step 1: Find the center O and radius of the given circle.</strong></p><p>Rewrite: $x^2 + y^2 - 10x - 6y + 30 = 0$ as $(x-5)^2 + (y-3)^2 = 4$</p><p>Center: $O = (5, 3)$, Radius: $r = 2$</p><p><strong>Step 2: Verify that A(7,3) and B(5,1) lie on the circle.</strong></p><p>For A(7,3): $(7-5)^2 + (3-3)^2 = 4$ ✓</p><p>For B(5,1): $(5-5)^2 + (1-3)^2 = 4$ ✓</p><p><strong>Step 3: Find the equation of chord AB.</strong></p><p>Slope of AB: $m = \frac{1-3}{5-7} = \frac{-2}{-2} = 1$</p><p>Line through A(7,3): $y - 3 = 1(x - 7)$ → $x - y - 4 = 0$</p><p><strong>Step 4: Write the family of circles S = 0.</strong></p><p>Family: $(x^2 + y^2 - 10x - 6y + 30) + \lambda(x - y - 4) = 0$</p><p>Expanding: $x^2 + y^2 - 10x - 6y + 30 + \lambda x - \lambda y - 4\lambda = 0$</p><p>Rearranging: $x^2 + y^2 + (-10 + \lambda)x + (-6 - \lambda)y + (30 - 4\lambda) = 0$</p><p><strong>Step 5: Find the smallest circle in the family.</strong></p><p>The smallest circle has AB as its diameter. Center is midpoint of AB: $M = \left(\frac{7+5}{2}, \frac{3+1}{2}\right) = (6, 2)$</p><p>Radius: $r = \frac{|AB|}{2} = \frac{\sqrt{(7-5)^2 + (3-1)^2}}{2} = \frac{\sqrt{8}}{2} = \sqrt{2}$</p><p><strong>Step 6: Write the equation of the smallest circle.</strong></p><p>$(x-6)^2 + (y-2)^2 = 2$</p><p>Expanding: $x^2 - 12x + 36 + y^2 - 4y + 4 = 2$</p><p>$x^2 + y^2 - 12x - 4y + 38 = 0$</p><p><strong>Step 7: Verify option (c).</strong></p><p>The equation $x^2 + y^2 - 12x - 4y + 38 = 0$ matches exactly with the given option (c).</p><p><strong>∴ Answer: c</strong></p>
Correct Answer: c