Question #b8a225ff - SAT Suite Question Bank

ID: b8a225ff

Modded SAT Question Bank by Abdullah Mallik

Circle A in the xy-plane has the equation ${(x + 5)}^{2} + {(y - 5)}^{2} = 4$ . Circle B has the same center as circle A. The radius of circle B is two times the radius of circle A. The equation defining circle B in the xy-plane is ${(x + 5)}^{2} + {(y - 5)}^{2} = k$ , where $k$ is a constant. What is the value of $k$ ?

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Correct Answer: 16

Rationale

The correct answer is $16$ . An equation of a circle in the xy-plane can be written as ${(x - t)}^{2} + {(y - u)}^{2} = r^{2}$ , where the center of the circle is $(t, u)$ , the radius of the circle is $r$ , and where $t$ , $u$ , and $r$ are constants. It’s given that the equation of circle A is ${(x + 5)}^{2} + {(y - 5)}^{2} = 4$ , which is equivalent to ${(x + 5)}^{2} + {(y - 5)}^{2} = 2^{2}$ . Therefore, the center of circle A is $left parenthesis negative 5 comma 5 right parenthesis$ and the radius of circle A is $2$ . It’s given that circle B has the same center as circle A and that the radius of circle B is two times the radius of circle A. Therefore, the center of circle B is $left parenthesis negative 5 comma 5 right parenthesis$ and the radius of circle B is $2 left parenthesis 2 right parenthesis$ , or $4$ . Substituting $negative 5$ for $t$ , $5$ for $u$ , and $4$ for $r$ into the equation ${(x - t)}^{2} + {(y - u)}^{2} = r^{2}$ yields ${(x + 5)}^{2} + {(y - 5)}^{2} = 4^{2}$ , which is equivalent to ${(x + 5)}^{2} + {(y - 5)}^{2} = 16$ . It follows that the equation of circle B in the xy-plane is ${(x + 5)}^{2} + {(y - 5)}^{2} = 16$ . Therefore, the value of $k$ is $16$ .

Question Difficulty: Hard