SOLVE STEP BY STEP IN DIGITAL FORMATFind the equation of the circle that passes through the intersections of the circles x² + y² - 6x + 4 = 0,x² + y² - 2 = 0 and is tangent to the line x + 3y - 14 = 0.

Answered: Find the equation of the circle that…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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SOLVE STEP BY STEP IN DIGITAL FORMAT
Find the equation of the circle that passes through the intersections of the circles x² + y² - 6x + 4 = 0,
x² + y² - 2 = 0 and is tangent to the line x + 3y - 14 = 0.

Expert Solution

Step 1: Finding the equation of the circle

The circle passes through the intersection of

$S_{1} = x^{2} + y^{2} - 6 x + 4 = 0 S_{2} = x^{2} + y^{2} - 2 = 0$

The intersection of the two circle is

Then the equation $S_{1} + λ S_{2} = 0$ , where $λ$ is the parameter, represents circle passing through the intersection of $S_{1}$ and $S_{2}$

$\Rightarrow (x^{2} + y^{2} - 6 x + 4) + λ (x^{2} + y^{2} - 2) = 0 \Rightarrow (λ + 1) (x^{2} + y^{2}) - 6 x + (4 - 2 λ) = 0 \Rightarrow x^{2} + y^{2} - \frac{6 x}{λ + 1} + \frac{4 - 2 λ}{λ + 1} = 0$

Then the center of the circle is $(\frac{3}{λ + 1}, 0)$

Form the general equation of circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ the radius is $\sqrt{f^{2} + g^{2} - c}$

$\Rightarrow R = \sqrt{{(\frac{3}{λ + 1})}^{2} - \frac{4 - 2 λ}{λ + 1}} = \frac{1}{λ + 1} \sqrt{9 - (4 - 2 λ) (λ + 1)} = \frac{1}{λ + 1} \sqrt{9 - (4 λ + 4 - 2 λ^{2} - 2 λ)} = \frac{1}{λ + 1} \sqrt{5 - 2 λ + 2 λ^{2}}$

We know that the distance of the point $(x_{0}, y_{0})$ from the straight line $a x + b + c = 0$ is $\frac{|a x_{0} + b y_{0} + c|}{\sqrt{a^{2} + b^{2}}}$

The distance of the tangent from the center of the circle is the radius of the circle

$\Rightarrow \frac{\sqrt{5 - 2 λ + 2 λ^{2}}}{λ + 1} = \frac{|\frac{3}{λ + 1} + 3 \cdot 0 - 14|}{\sqrt{1^{2} + 3^{2}}} \Rightarrow {(\frac{\sqrt{5 - 2 λ + 2 λ^{2}}}{λ + 1})}^{2} = {(\frac{\frac{3}{λ + 1} - 14}{\sqrt{10}})}^{2} \Rightarrow \frac{5 - 2 λ + 2 λ^{2}}{λ^{2} + 2 λ + 1} = \frac{1}{10} {(\frac{3 - 14 (λ + 1)}{λ + 1})}^{2} \Rightarrow 50 - 20 λ + 20 λ^{2} = {(14 λ + 11)}^{2} \Rightarrow 50 - 20 λ + 20 λ^{2} = 196 λ^{2} + 308 λ + 121 \Rightarrow 176 λ^{2} + 328 λ + 71 = 0 \Rightarrow λ = - \frac{1}{4}, - 1.61$

Therefore the equation of the circle is

$x^{2} + y^{2} - \frac{6 x}{λ + 1} + \frac{4 - 2 λ}{λ + 1} = 0 \Rightarrow x^{2} + y^{2} - \frac{6 x}{- 0.25 + 1} + \frac{4 + 0.5}{- 0.25 + 1} = 0 \Rightarrow x^{2} + y^{2} - 8 x + 6 = 0$