Circles
Equation of circle
Grade 11
Question:
<p>The lines \(2x - 3y = 5\) and \(3x - 4y = 7\) are diameters of a circle having area as 154 sq. units. Then the equation of the circle is</p>
<p>\(x^2 + y^2 + 2x - 2y - 62 = 0\)</p>
<p>\(x^2 + y^2 + 2x - 2y - 47 = 0\)</p>
<p>\(x^2 + y^2 - 2x + 2y - 47 = 0\)</p>
<p>\(x^2 + y^2 - 2x + 2y - 62 = 0\)</p>
Step-by-Step Solution
Key Concept: Since two lines are diameters of a circle, their intersection point is the center. Use the area to find the radius, then construct the circle equation using center and radius.
<p><strong>Step 1: Find the center by solving the two diameter equations.</strong></p><p>From 2x - 3y = 5 and 3x - 4y = 7:</p><p>Multiply first equation by 3: 6x - 9y = 15</p><p>Multiply second equation by 2: 6x - 8y = 14</p><p>Subtract: -y = 1, so y = -1</p><p>Substitute back: 2x - 3(-1) = 5 → 2x = 2 → x = 1</p><p><strong>Center: (1, -1)</strong></p><p><strong>Step 2: Find the radius using the area.</strong></p><p>Area = πr² = 154</p><p>r² = 154/π = 154/(22/7) = 154 × 7/22 = 49</p><p>r = 7</p><p><strong>Step 3: Write the circle equation.</strong></p><p>With center (1, -1) and radius 7:</p><p>(x - 1)² + (y + 1)² = 49</p><p>Expanding: x² + y² - 2x + 2y + 1 + 1 = 49</p><p><strong>x² + y² - 2x + 2y - 47 = 0</strong></p><p>∴ Answer: C</p>
Correct Answer: C