How do you write the equation of the circle a center at (5, -1) and containing the point (4,-5)?

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How do you write the equation of the circle a center at (5, -1) and containing the point (4,-5)?

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Answer 1 · 2018-04-25T14:12:45

${(x - 5)}^{2} + {(y + 1)}^{2} = 37$ $(x-5)^2 + (y+1)^2 =37$

Explanation:

Find the distance to (4,-5) from (5,-1) first, which is $\sqrt{{(5 - 4)}^{2} + {(- 1 - 5)}^{2}}$ $sqrt((5-4)^2+(-1-5)^2$ = $\sqrt{1 + 36}$ $sqrt(1 + 36)$ = $\sqrt{37}$ $sqrt37$ hence the radius is $\sqrt{37}$ $sqrt37$ . Use the equation ${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $(x-a)^2 + (y-b)^2 = r^2$ where a is the x co-ordinate of the centre of the circle and b is the y co-ordinate. r is the radius.

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How do you write the equation of the circle a center at (5, -1) and containing the point (4,-5)?

1 Answer

Explanation:

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