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

Answer 1

#(x-5)^2 + (y+1)^2 =37#

Find the distance to (4,-5) from (5,-1) first, which is#sqrt((5-4)^2+(-1-5)^2# = #sqrt(1 + 36)# = #sqrt37# hence the radius is #sqrt37#. Use the equation #(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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Answer 2

James Atkins

The equation of a circle with center ( (h, k) ) and radius ( r ) is given by:

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

For the circle with center at ( (5, -1) ) and containing the point ( (4, -5) ), we substitute the center coordinates into ( (h, k) ) and the given point into ( (x, y) ):

[ (x - 5)^2 + (y + 1)^2 = r^2 ]

Now, we need to find the radius ( r ). Since the given point ( (4, -5) ) lies on the circle, it satisfies the equation of the circle:

[ (4 - 5)^2 + (-5 + 1)^2 = r^2 ]

[ (-1)^2 + (-4)^2 = r^2 ]

[ 1 + 16 = r^2 ]

[ 17 = r^2 ]

So, the equation of the circle is:

[ (x - 5)^2 + (y + 1)^2 = 17 ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.