Question Detail

MTH 101 MTH 101 Past Questions Objective Question
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Question

Find the equation of a circle whose center is at the point (-1,6) and passes through the point (2,3).
Options
A x² + y² + 2x - 12y + 19 = 0
Correct Answer
B x² + y² + 12x - 2y + 19 = 0
C x² + y² + 3x - 4y + 19 = 0
D x² + y² + 2x - 12y + 9 = 0
Correct Answer

Option A is the correct answer.

Detailed Explanation

Center (h,k) = (-1,6). Radius r = distance from center to (2,3) = √[(2-(-1))² + (3-6)²] = √[3² + (-3)²] = √(9+9) = √18 = 3√2. Equation: (x+1)² + (y-6)² = 18 → x² + 2x + 1 + y² - 12y + 36 = 18 → x² + y² + 2x - 12y + 37 - 18 = 0 → x² + y² + 2x - 12y + 19 = 0.

Hint

Use the standard form (x-h)² + (y-k)² = r².

Question Info

Course: MTH 101
Topic: MTH 101 Past Questions
Type: Objective
Difficulty: Not specified
Question 439 of 696
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