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What is the general form of the equation of a circle with the center at (-2,1) and passing through (-4,1)

User Ysh
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1 Answer

7 votes

Answer:


\large\boxed{x^2+y^2+4x-2y+1=0}

Explanation:

The standard form of an equation of a circle:


(x-h)^2+(y-k)^2=r^2

(h, k) - center

r - radius

The general form of an equation of a circle:


x^2+y^2+Dx+Ey+F=0

We have the center (-2, 1). Substitute to the equation in the standard form:


(x-(-2))^2+(y-1)^2=r^2\\\\(x+2)^2+(y-1)^2=r^2

Put thr coordinates of the point (-4, 1) to the equation and calculate a radius:


(-4+2)^2+(1-1)^2=r^2\\\\r^2=(-2)^2+0^2\\\\r^2=4

Therefore we have the equation:


(x+2)^2+(y-1)^2=4

Convert to the general form.

Use
(a\pm b)^2=a^2\pm 2ab+b^2


(x+2)^2+(y-1)^2=4\\\\x^2+2(x)(2)+2^2+y^2-2(y)(1)+1^2=4\\\\x^2+4x+4+y^2-2y+1=4\qquad\text{subtract 4 from both sides}\\\\x^2+y^2+4x-2y+1=0

User Oli
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