Questions: The circle above is tangent to both the horizontal and vertical lines. It also touches the corner of the .5 by .75 rectangle. Solve for the radius of the circle. A r ≈ 5.27 B r ≈ 8.25 C r ≈ 15.71 D r ≈ 2.57 E none of these

Solution Steps

The circle is tangent to both the horizontal and vertical lines and touches the corner of a 0.5 by 0.75 rectangle. We need to find the radius \( r \) of the circle.

Since the circle is tangent to both axes, the center of the circle is at \((r, r)\). The circle also touches the corner of the rectangle at \((0.5, 0.75)\).

The distance from the center of the circle \((r, r)\) to the point \((0.5, 0.75)\) is equal to the radius \( r \). Using the distance formula: \[ \sqrt{(r - 0.5)^2 + (r - 0.75)^2} = r \]

Square both sides to remove the square root: \[ (r - 0.5)^2 + (r - 0.75)^2 = r^2 \]

Expand the squares: \[ (r^2 - r + 0.25) + (r^2 - 1.5r + 0.5625) = r^2 \] Combine like terms: \[ 2r^2 - 2.5r + 0.8125 = r^2 \]

Move all terms to one side: \[ r^2 - 2.5r + 0.8125 = 0 \] Solve the quadratic equation using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ r = \frac{2.5 \pm \sqrt{(2.5)^2 - 4 \cdot 1 \cdot 0.8125}}{2 \cdot 1} \] \[ r = \frac{2.5 \pm \sqrt{6.25 - 3.25}}{2} \] \[ r = \frac{2.5 \pm \sqrt{3}}{2} \] \[ r = \frac{2.5 \pm 1.732}{2} \]

\[ r = \frac{2.5 + 1.732}{2} \approx 2.116 \] \[ r = \frac{2.5 - 1.732}{2} \approx 0.384 \]

Since the radius must be positive and fit the given options, we check the closest value: \[ r \approx 2.116 \]

Final Answer