I'm working midpoints, distance, circles and standard form and general form. I have some questions with my answers to the first two-I'm stuck on #3
1. What is the distance between points (0,0) and (-3,5)
My answer: sqrt34
2.What is distance between (3,1) and (10,8)
My answer is 7sqrt2
3. I need the equation in general form (x^2 + y^2 + Dx + Ey +F = 0 for a circle whose center is (-3,-6) and passes through point (7,-3)
First I found radius which is distance between the two points = sqrt109
then it is ( ^2 + (y+6)^2 = 109
and then I'm lost please helpx+3)
1. correct
2. correct
3. correct: (x+3)^2 + (y+6)^2 = 109
Expand all that to get the general form.
Ah. you're lost?
(x+3)^2 + (y+6)^2 = 109
x^2 + 6x + 9 + y^2 + 12y + 36 = 109
x^2 + y^2 + 6x + 12y - 64 = 0
Thank you
To find the equation of a circle in general form given its center and a point on the circle, you can follow these steps:
1. Find the radius:
- You've correctly found the distance between the center (-3, -6) and the point on the circle (7, -3), which is √109.
2. Write the equation using the distance formula:
- The general equation for a circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r represents the radius.
- Substitute the center (-3, -6) into (h, k) and the radius √109 into r. The equation becomes:
(x - (-3))^2 + (y - (-6))^2 = (√109)^2
Simplifying, we get:
(x + 3)^2 + (y + 6)^2 = 109
3. Convert the equation to general form:
- Expand the equation by squaring the binomials:
(x + 3)(x + 3) + (y + 6)(y + 6) = 109
Simplify further:
x^2 + 6x + 9 + y^2 + 12y + 36 = 109
Rearrange the terms and bring 109 to the right side:
x^2 + y^2 + 6x + 12y + 45 = 0
So, the equation in general form for the circle with the center (-3, -6) and passing through (7, -3) is:
x^2 + y^2 + 6x + 12y + 45 = 0