/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 733 Find the equation of the circle ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 42: Problem 733

Find the equation of the circle that goes through the points \((1,2)\) and \((3,4)\) and has radius \(\mathrm{a}=2\).

Short Answer

Expert verified
The equation of the circle that goes through points \((1,2)\) and \((3,4)\) and has radius \(2\) is \((x-1)^2 + y^2 = 4\).

Step by step solution

01

Write down the distance formula

The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
02

Apply the distance formula for the given points and center\((h, k)\)

Using the distance formula for points \((1, 2)\), \((3,4)\) and the circle's center \((h, k)\), we get two equations: \[ 2=\sqrt{(h-1)^2+(k-2)^2} \] \[ 2=\sqrt{(h-3)^2+(k-4)^2} \] Square both sides to get rid of the square roots: \[ 4=(h-1)^2+(k-2)^2 \] \[ 4=(h-3)^2+(k-4)^2 \]
03

Solve the system of equations for \(h\) and \(k\)

Start by subtracting the first equation from the second equation: \[ 0=(h-3)^2+(k-4)^2-(h-1)^2-(k-2)^2 \] Now, simplify and rewrite the equation: \[ 0=4(h-2)+4(k-3) \] Next, simplify again: \[ 0=h-2+k-3 \] Now, we have a simpler linear equation. Solve for one of the variables, say \(k\): \[ k=h-1 \] Now, plug this expression into either of the original squared equations. We'll use the first one: \[ 4=(h-1)^2+[(h-1)-2]^2 \]
04

Solve for \(h\)

Simplify the equation and solve for \(h\): \[ 4 = (h-1)^2+(h-3)^2 \] Expand the squares: \[ 4 = h^2-2h+1+h^2-6h+9 \] Combine like terms: \[ 4 = 2h^2-8h+10 \] Divide by 2 to simplify: \[ 2 = h^2-4h+5 \] Set it equal to zero: \[ 0 = h^2-4h+3 \] Factor: \[ 0 = (h-1)(h-3) \] So, \(h = 1\) or \(h = 3\).
05

Solve for \(k\) for each value of \(h\)

Compute \(k\) using \(k=h-1\)for both \(h=1\) and \(h=3\). \[ k_1 = 1-1 = 0 \] \[ k_2 = 3-1 = 2 \]
06

Verify which center works

Plug both \((h_1, k_1) = (1, 0)\) and \((h_2, k_2) = (3, 2)\) into the original radius equation. Since the radius should be 2, we will pick the one that satisfies the condition. For \((1, 0)\): \[ 2=\sqrt{(1-1)^2+(2-0)^2}=2 \] For \((3, 2)\): \[ 2=\sqrt{(3-1)^2+(4-2)^2}=\sqrt{8} \neq 2 \] So, the center \((h, k)\) is \((1, 0)\).
07

Write the equation of the circle

Finally, we write the equation of the circle using the general formula \((x-h)^2 + (y-k)^2 = a^2\), with the center \((1, 0)\) and radius \(2\): \[ (x-1)^2 + y^2 = 4 \] Thus, the equation of the circle that goes through points \((1,2)\) and \((3,4)\) and has radius \(2\) is: \[ (x-1)^2 + y^2 = 4 \]

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

A single-lane highway must pass under a series of bridges. It is proposed that the bridges be shaped as semi-ellipses with the height equal to the width. The builder feels he must allow room for a 6 foot wide, 12 foot high truck to pass under it. What is the lowest bridge that can be built to serve this purpose.

Write the equation of the circle with center \(\mathrm{C}\) at the origin and with radius 7 .

Given that two circles \(\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{D}_{1} \mathrm{x}+\mathrm{E}_{1} \mathrm{y}+\mathrm{F}_{1}=0\) and \(\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{D}_{2} \mathrm{x}+\mathrm{E}_{2} \mathrm{y}+\mathrm{F}_{2}=0\) intersect at two points, show that the equation for the line determined by the points of intersection is \(\left(\mathrm{D}_{1}-\mathrm{D}_{2}\right) \mathrm{x}+\left(\mathrm{E}_{1}-\mathrm{E}_{2}\right)+\left(\mathrm{F}_{1}-\mathrm{F}_{2}\right)=0\).

Consider a point \(P_{1}\left(x_{1}, y_{1}\right)\) on the ellipse \(b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}\) A tangent to the ellipse at \(p_{1}\) is a line through \(p_{1}\) with no other point on the ellipse. Prove that if \(y_{1} \neq 0\), there is a tangent at \(\mathrm{p}_{1}\), its slope is \(\mathrm{m}=\left(-\mathrm{b}^{2} \mathrm{x}_{1}\right) /\left(\mathrm{a}^{2} \mathrm{y}_{1}\right)\) and its equation can be put in the form \(\mathrm{x}_{1} \mathrm{x} / \mathrm{a}^{2}+\mathrm{y}_{1} \mathrm{y} / \mathrm{b}^{2}=1\).

Write the equation of the circle with its center at the origin and with radius of length \(3 .\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Statistics

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.