Circle And System Of Circles Question 231
Question: A circle passes through the points (0, 0) and (4, 0) and touches the line y = 2. The equation of the circle is
Options:
A) $x^{2} + y^{2} - 4x - 2y = 0$
B) $x^{2} + y^{2} - 4x + 2y = 0$
C) $x^{2} + y^{2} + 4x - 2y = 0$
D) $x^{2} + y^{2} + 4x + 2y = 0$
Show Answer
Answer:
Correct Answer: A
Solution:
The circle passes through the origin (0, 0) and (4, 0), so the center of the circle must be at (2, 0). Since the circle touches the line y = 2, the radius of the circle must be 2. Therefore, the equation of the circle is $(x - 2)^{2} + (y - 0)^{2} = 2^{2}$, which simplifies to $x^{2} + y^{2} - 4x - 2y = 0$. So, the correct answer is Option A: $x^{2} + y^{2} - 4x - 2y = 0$. This follows from the standard form of the equation of a circle and the properties of a circle.
BETA
