Circles
Equation of circle given diameters
Grade 11
Question:
<p>If the lines \(3x - 4y - 7 = 0\) and \(2x - 3y - 5 = 0\) are two diameters of a circle of area \(49\pi\) square units, the equation of the circle is</p>
<p>\(x^2 + y^2 + 2x - 2y - 47 = 0\)</p>
<p>\(x^2 + y^2 + 2x - 2y - 62 = 0\)</p>
<p>\(x^2 + y^2 - 2x + 2y - 62 = 0\)</p>
<p>\(x^2 + y^2 - 2x + 2y - 47 = 0\)</p>
Step-by-Step Solution
Key Concept: The center of a circle lies at the intersection of any two diameters. Find this intersection point, then use the given area to determine the radius and write the circle equation.
<p><strong>Step 1:</strong> Find the center by solving the system of diameter equations:</p><p>3x - 4y - 7 = 0 ... (1)<br>2x - 3y - 5 = 0 ... (2)</p><p>Multiply (1) by 3: 9x - 12y - 21 = 0<br>Multiply (2) by 4: 8x - 12y - 20 = 0<br>Subtract: x - 1 = 0 ⟹ x = 1</p><p>Substitute x = 1 in (1): 3(1) - 4y - 7 = 0 ⟹ -4y = 4 ⟹ y = -1</p><p><strong>Step 2:</strong> Find radius from area:</p><p>Area = 49π = πr²<br>r² = 49 ⟹ r = 7</p><p><strong>Step 3:</strong> Write the circle equation with center (1, -1) and radius 7:</p><p>(x - 1)² + (y + 1)² = 49</p><p>Expanding: x² - 2x + 1 + y² + 2y + 1 = 49<br>x² + y² - 2x + 2y - 47 = 0</p><p>∴ Answer: D</p>
Correct Answer: D