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What is the equation of a circle with center (-3, -1) that contains the point (1, 2)?

User Dean Seo
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2 Answers

1 vote

Answer:

(x + 3)^2 + (y + 1^)2 = 25

Explanation:

The distance formula gives the length of the radius:

r = √(-3-1)^2 + (- 1-2)^2 = √25 = 5

Substituting h = -3, k = -1, and r = 5 into the standard form (x - h)2 + (y - k)2 = r2, you get (x + 3)2 + (y + 1)2 = 52 = 25.

edmentum :)

User Rocky Qi
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7.2k points
6 votes
so, we know the center is at -3,-1, ok

hmmm what's the radius anyway? well, we know that there's a point at 1,2 that is on the circle's path...hmmmm what's the distance from the center to that point? well, is the radius, let's check then.


\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -3}}\quad ,&{{ -1}})\quad % (c,d) &({{ 1}}\quad ,&{{ 2}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ r=√([1-(-3)]^2+[2-(-1)]^2)\implies r=√((1+3)^2+(2+1)^2) \\\\\\ r=√(16+9)\implies r=√(25)\implies r=5

so, what's the equation of a circle with center at -3, -1 and a radius of 5?


\bf \textit{equation of a circle}\\\\ (x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2 \qquad \begin{array}{lllll} center\ (&{{ h}},&{{\quad k}})\qquad radius=&{{ r}}\\ &-3&-1&5 \end{array} \\\\\\\ [x-(-3)]^2+[y-(-1)]^2=5^2\implies (x+3)^2+(y+1)^2=25
User ZeroCho
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8.0k points