What is the standard form of the equation of a circle with center of a circle is at (-15,32) and passes through the point (-18,21)?

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What is the standard form of the equation of a circle with center of a circle is at (-15,32) and passes through the point (-18,21)?

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Answer 1 · 2015-11-14T20:57:01

${(x + 15)}^{2} + {(y - 32)}^{2} = 130$ $(x+15)^2+(y-32)^2=130$

Explanation:

The standard form of a circle centred at (a,b) and having radius r is ${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $(x-a)^2+(y-b)^2=r^2$ .

So in this case we have the centre, but we need to find the radius and can do so by finding the distance from the centre to the point given :

$d ((- 15, 32); (- 18, 21)) = \sqrt{{(- 18 - (- 15))}^{2} + {(21 - 32)}^{2}} = \sqrt{130}$ $d((-15,32);(-18,21))=sqrt((-18-(-15))^2+(21-32)^2)=sqrt130$

Therefore the equation of the circle is

${(x + 15)}^{2} + {(y - 32)}^{2} = 130$ $(x+15)^2+(y-32)^2=130$

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What is the standard form of the equation of a circle with center of a circle is at (-15,32) and passes through the point (-18,21)?

1 Answer

Explanation:

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