A circle in the xy-plane has its center at (-7, 3) and has a radius of 9. An equation of this circle is x^2 + y^2 + ax + by + c = 0, where a, b, and c are constants. What is the value of c?

A circle in the xy-plane has its center at (-7, 3) and has a radius of 9. An equation of... | SAT Circles Practice

Answer: $-23$

Explanation:

$1$ . Write the equation of the circle in standard (center-radius) form:

- The standard form is $(x-h)^2 + (y-k)^2 = r^2$ , where $(h, k)$ is the center and r is the radius.

- Substitute the given center $(-7, 3)$ and radius $9$ :

$(x - (-7))^2 + (y - 3)^2 = 9^2$

$(x + 7)^2 + (y - 3)^2 = 81$

$2$ . Expand the equation to the general form $x^2 + y^2 + ax + by + c = 0$ :

- Expand the squared binomials:

$(x^2 + 14x + 49) + (y^2 - 6y + 9) = 81$

- Group the terms and move the constant from the right side to the left:

$x^2 + y^2 + 14x - 6y + 49 + 9 - 81 = 0$

$3$ . Simplify to find the value of c:

- $x^2 + y^2 + 14x - 6y - 23 = 0$

- Comparing this to the general form, the constant term c is $-23$ .

$4$ . Conclusion:

- The value of c is $-23$ .