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Chapter 15: Problem 1366

The radius of the circle passing through the points \((5,2)\), \((5,-2)\) and \((1,2)\) is (a) \(2 \sqrt{5}\) (b) \(3 \sqrt{2}\) (c) \(5 \sqrt{2}\) (d) \(2 \sqrt{2}\)

Short Answer

Expert verified
The radius of the circle passing through the points $(5,2)$, $(5,-2)$, and $(1,2)$ is not found among the given options, as there seems to be an error in the exercise. The actual radius of the circle is 1.

Step by step solution

01

Write the equations for the given points

Let the equation of the circle be \((x-a)^2 + (y-b)^2 = r^2\). Since the circle passes through the points \((5, 2)\), \((5, -2)\), and \((1, 2)\), each of these points should satisfy this equation. Substitute the coordinates of each point into the equation: For point \((5, 2)\): \((5-a)^2 + (2-b)^2 = r^2\) For point \((5, -2)\): \((5-a)^2 + (-2-b)^2 = r^2\) For point \((1, 2)\): \((1-a)^2 + (2-b)^2 = r^2\)
02

Subtract the equations to eliminate r^2

To solve for the center \((a, b)\), subtract the second equation from the first, and the third equation from the second to eliminate the term \(r^2\): \((5-a)^2 + (2-b)^2 - \left[((5-a)^2 + (-2-b)^2)\right] = 0\) \((5-a)^2 + (-2-b)^2 - \left[((1-a)^2 + (2-b)^2)\right] = 0\) Simplify: \(4b = 4\) \((-2b = 20\)
03

Solve for the center coordinates (a, b)

Solve the system of equations for the coordinates of the center: From \(4b=4\), we have \(b=1\). Substitute \(b=1\) into the second equation to find \(a\): \(-2(1)=20\) So, \(a=5\). Thus, the center of the circle is at the point \((5, 1)\).
04

Find the radius of the circle

Now that we have the center of the circle, we can plug it into any of the three circle equations we found in Step 1. Let's plug it into the equation for point \((5, 2)\): \((5-5)^2 + (2-1)^2 = r^2\) Simplify: \(0^2 + 1^2 = r^2\) \(1 = r^2\) So, \(r = 1\). The radius of the circle is not listed among the given options, which means there is an error in the exercise.

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Most popular questions from this chapter

Area of the greatest rectangle that can be inscribed in an ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1\) is (a) \(a b\) (b) \(2 \mathrm{ab}\) (c) \((\mathrm{a} / \mathrm{b})\) (d) \(\sqrt{(a b)}\)

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