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Please answer quickly!! It's just a pre-cal circles equation

Please answer quickly!! It's just a pre-cal circles equation-example-1
User Odlp
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Answer: the equation of the circle in standard form is (x + 3)^2 + (y + 11)^2 = 121.

Step-by-step explanation: To find the center of the circle, we can substitute the coordinates of the point where the circle is tangent to the x-axis, which is (-3,0), into the equation of the line y = 4x + 1.

Substituting -3 for x, we get:

y = 4(-3) + 1

y = -12 + 1

y = -11

Therefore, the center of the circle is (-3, -11).

To find the radius, we need to measure the distance from the center of the circle to any point on the circle. Since the circle is tangent to the x-axis at (-3,0), the distance from the center to this point is the radius.

Using the distance formula, we can calculate the radius:

r = √[(x2 - x1)^2 + (y2 - y1)^2]

r = √[(-3 - (-3))^2 + (0 - (-11))^2]

r = √[(0)^2 + (11)^2]

r = √(0 + 121)

r = √121

r = 11

Therefore, the radius of the circle is 11.

Now we have the center (-3, -11) and the radius 11. We can plug these values into the standard equation of a circle to get the equation in standard form:

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

(x - (-3))^2 + (y - (-11))^2 = 11^2

(x + 3)^2 + (y + 11)^2 = 121

User Serg Kryvonos
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