Sample Question 10:

In the \(xy\)-plane, a circle of radius \(4\) with center on the positive \(x\)-axis is tangent to the \(y\)-axis at the origin, and a circle with radius \(10\) with center on the positive \(y\)-axis is tangent to the \(x\)-axis at the origin. What is the slope of the line passing through the two points at which these circles intersect?

\(\textbf{(A)}\ \dfrac{2}{7} \qquad\textbf{(B)}\ \dfrac{3}{7} \qquad\textbf{(C)}\ \dfrac{2}{\sqrt{29}} \qquad\textbf{(D)}\ \dfrac{1}{\sqrt{29}} \qquad\textbf{(E)}\ \dfrac{2}{5}\)

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Answer Keys

Question 10: E

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Solutions

Question 10
Step 1: Identify the centers of the two circles. The center of the first circle is at (4,0) and the center of the second circle is at (0,10).

Step 2: Calculate the slope of the line that passes through these two centers. The slope is \(-\dfrac{10}{4} = -\dfrac{5}{2}\).

Step 3: Keep in mind that this line is the perpendicular bisector of the line that passes through the two intersecting points of the circles.

Step 4: To find the slope of the line passing through the points of intersection of the two circles, take the negative reciprocal of the slope of the line passing through the centers. The slope of this line is \(-1 \div -\dfrac{5}{2} = \dfrac{2}{5}\). Hence, the answer is \(\textbf{(E)}\ \dfrac{2}{5}\).