Circles
Equation of circle
Grade 11
Question:
<p>If the lines \(2x + 3y + 1 = 0\) and \(3x - y - 4 = 0\) lie along diameters of a circle of circumference \(10\pi\), then the equation of the circle is</p>
<p>\(x^2 + y^2 - 2x + 2y - 23 = 0\)</p>
<p>\(x^2 + y^2 - 2x - 2y - 23 = 0\)</p>
<p>\(x^2 + y^2 + 2x + 2y - 23 = 0\)</p>
<p>\(x^2 + y^2 + 2x - 2y - 23 = 0\)</p>
Step-by-Step Solution
Key Concept: The intersection point of two diameter lines is the center of the circle. The circumference gives the radius, which determines the circle's equation completely.
<p><strong>Step 1: Find the center by solving the two diameter equations simultaneously.</strong></p><p>Lines: 2x + 3y + 1 = 0 and 3x - y - 4 = 0</p><p>From second equation: y = 3x - 4</p><p>Substitute into first: 2x + 3(3x - 4) + 1 = 0</p><p>2x + 9x - 12 + 1 = 0</p><p>11x = 11 ⟹ x = 1</p><p>y = 3(1) - 4 = -1</p><p><strong>Center: (1, -1)</strong></p><p><strong>Step 2: Find radius from circumference.</strong></p><p>Circumference = 10π</p><p>2πr = 10π ⟹ r = 5</p><p>r² = 25</p><p><strong>Step 3: Write the circle equation.</strong></p><p>Standard form: (x - h)² + (y - k)² = r²</p><p>(x - 1)² + (y + 1)² = 25</p><p>Expanding: x² - 2x + 1 + y² + 2y + 1 = 25</p><p><strong>x² + y² - 2x + 2y - 23 = 0</strong></p><p>∴ Answer: B</p>
Correct Answer: B