How do you write an equation with center is (4,-1) and solution point is (1,4)?

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Answer 1 · 2016-12-25T02:13:20

I am going to assume that you want the equation of a circle. Please see the explanation.

The standard Cartesian form for the equation of a circle is:

$(x - h)^2 + (y - k)^2 = r^2" [1]"$

where x and y are any point, $(x, y)$ , on the circle, h and k are the center point, $(h, k)$ , and r is the radius.

Substitute 4 for h and -1 for k into equation [1]:

$(x - 4)^2 + (y - -1)^2 = r^2" [2]"$

To find the value of r, substitute 1 for x and 4 for y into equation [2], then solve for r:

$(1 - 4)^2 + (4 - -1)^2 = r^2$

$3^2 + 5^2 = r^2$

$r^2 = 36$

$r = 6$

Substitute 6 for r into equation [2]:

$(x - 4)^2 + (y - -1)^2 = 6^2" [3]"$

Equation [3] represents the desired circle.