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In $20-27$ : a. Write each equation in center-radius form. b. Find the coordinates of the center. . Find the radius of the circle.$$x^{2}+y^{2}+2 x-4 y+1=0$$

(a) $(x+1)^{2}+(y-2)^{2}=4$(b) $(-1,2)$(c) $2$

Algebra

Chapter 4

RELATIONS AND FUNCTIONS

Section 9

Circles

An Introduction to Geometry

Functions

Linear Functions

Polynomials

Campbell University

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for this problem, we're going to find the standard centrist and reform of the given equation of the circle. And then from that find the center and the radius. So for this one will need Thio complete the square and work to find the components X and Y on the left side. And then we'll get the car one of the constant that we find along with that over to the right side. So we have two bunch are ex together and bunch or wives together. So we'll have X group was two x plus some value. Ah, blank Plus why squared minus four? Why plus some value blank. Ah is equal to minus one because, well, scoot the constants of the right. So we'll add one to complete the square for X square plus ticks plus one because to over +12 over two is one and one squared is one. Ah, for this one will do minus four. Divided by two is minus two minus two. Whole squared is for So that's what we put there. So on the right side will have to add the same values we add one and we add for so because of that yet that the standards under standard equation of the circle his expose expose one whole squared plus why minus two whole squared is equal to four. So from that will get the center age comma K. So the center is equal to minus one comma, too. And the radius is equal. Well, we have our square from the equation is equal to four. If we take the square root of both sides will get the square root of four, which is equal to So that's all the equation we can. That's all the information we can describe from the equation.

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