Log out Go to App
Go to App

Learning Materials

Features

Discover

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If \(P\) is a point on an ellipse \(5 x^{2}+4 y^{2}=80\) whose foci are \(S\) and \(\mathrm{S}^{\prime}\). Then \(\mathrm{PS}+\mathrm{PS}^{\prime}=\ldots \ldots \ldots\) (a) \(4 \sqrt{5}\) (b) 4 (c) 8 (d) 10

If the line \(\mathrm{x}-1=0\) is the directrix of the parabola \(\mathrm{y}^{2}-\mathrm{kx}+8=0\) then one of the values of \(\mathrm{k}\) is (a) 4 (b) \((1 / 8)\) (c) \((1 / 4)\) (d) 8

Four distinct points \((1,0),(0,1),(0,0)\) and \((2 \mathrm{a}, 3 \mathrm{a})\) lie on a circle for (a) only one value of \(\mathrm{a} \in(0,1)\) (b) \(\mathrm{a}>2\) (c) \(\mathrm{a}<0\) (d) \(a \in(1,2)\)

Tangents are drawn to the ellipse \(\left(\mathrm{x}^{2} / 9\right)+\left(\mathrm{y}^{2} / 5\right)=1\) at ends of latus recturm line. The area of quadrilateral so formed is \(\ldots \ldots \ldots\) (a) \((27 / 4)\) (b) \((27 / 55)\) (c) 27 (d) \((27 / 2)\)

The equation of circle touching the axis of \(\mathrm{y}\) at a distance \(+4\) from the origin and cutoff an intercept 6 from the axis of \(\mathrm{x}\) is \(\ldots \ldots \ldots\) (a) \(x^{2}+y^{2}-10 x-8 y+16=0\) (b) \(x^{2}+y^{2}+10 x-8 y+16=0\) (c) \(x^{2}+y^{2}-10 x+8 y+16=0\) (d) none of these
See all solutions

Recommended explanations on English Textbooks

Textual Analysis

Read Explanation

Language Analysis

Read Explanation

Discourse

Read Explanation

Rhetorical Analysis Essay

Read Explanation

Semiotics

Read Explanation

Research and Composition

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free