Question #ebbf23ae - SAT Suite Question Bank

ID: ebbf23ae

Modded SAT Question Bank by Abdullah Mallik

A circle in the xy-plane has a diameter with endpoints $left parenthesis 2 comma 4 right parenthesis$ and $left parenthesis 2 comma 14 right parenthesis$ . An equation of this circle is ${(x - 2)}^{2} + {(y - 9)}^{2} = r^{2}$ , where $r$ is a positive constant. What is the value of $r$ ?

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Correct Answer: 5

Rationale

The correct answer is $5$ . The standard form of an equation of a circle in the xy-plane is ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ , where $h$ , $k$ , and $r$ are constants, the coordinates of the center of the circle are $(h, k)$ , and the length of the radius of the circle is $r$ . It′s given that an equation of the circle is ${(x - 2)}^{2} + {(y - 9)}^{2} = r^{2}$ . Therefore, the center of this circle is $left parenthesis 2 comma 9 right parenthesis$ . It’s given that the endpoints of a diameter of the circle are $left parenthesis 2 comma 4 right parenthesis$ and $left parenthesis 2 comma 14 right parenthesis$ . The length of the radius is the distance from the center of the circle to an endpoint of a diameter of the circle, which can be found using the distance formula, $\sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}}$ . Substituting the center of the circle $left parenthesis 2 comma 9 right parenthesis$ and one endpoint of the diameter $left parenthesis 2 comma 4 right parenthesis$ in this formula gives a distance of $\sqrt{{(2 - 2)}^{2} + {(9 - 4)}^{2}}$ , or $StartRoot 0 squared plus 5 squared EndRoot$ , which is equivalent to $5$ . Since the distance from the center of the circle to an endpoint of a diameter is $5$ , the value of $r$ is $5$ .

Question Difficulty: Hard