Question

What will be the equation of the circle whose center is \( (1, 2) \) and which passes through the point \( (4, 6) \)?

• A. \( x^2 + y^2 - 2x - 4y + 1 = 0 \)
• B. \( x^2 + y^2 + 2x + 4y + 1 = 0 \)
• C. \( x^2 + y^2 - 2x - 4y - 1 = 0 \)
• D. \( x^2 + y^2 + 2x + 4y - 1 = 0 \)

Hint:

To find the equation of a circle, first use the distance formula to find the radius, and then substitute into the standard circle equation.

Answer 1

The Correct Option is A

For a circle with center \( (h, k) \) and radius \( r \), the equation is: \[ (x - h)^2 + (y - k)^2 = r^2 \] We are given the center \( (h, k) = (1, 2) \). To find the radius, use the distance formula between the center and the point \( (4, 6) \): \[ r = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \] Thus, the equation of the circle is: \[ (x - 1)^2 + (y - 2)^2 = 25 \] Expanding: \[ (x^2 - 2x + 1) + (y^2 - 4y + 4) = 25 \] \[ x^2 + y^2 - 2x - 4y + 5 = 25 \] \[ x^2 + y^2 - 2x - 4y + 1 = 0 \] Thus, the equation of the circle is \( x^2 + y^2 - 2x - 4y + 1 = 0 \).