Write the standard form of the equation for the circle that passes through the points (2, 6), (-4, 6), and (-4, -2). Then identify the center and radius.
So far I got
2D+6E+F+40
-4D+6E+F+52
-4D-2E+F-20
IS this correct> and how do I solve it?
The standard form of a circle centred at (x0,y0) with a radius of r is given by:
(x-x0)²+(y-y0)²=r² ...(1)
method 1: algebraic
Given the three points, we can substitute (x1,y1) in equation to obtain an equation in x0,y0 and r.
Substituting the other two points will give two more equations which should enable us to solve for x0, y0 and r.
method 2: geometric
Take any two of the three points as a chord of the circle. Find the perpendicular bisector of the segment, L1. Take the third point together with one of the two previous points and make another perpendicular bisector, L2.
The intersection of L1 and L2 will be the centre O (x0,y0) of the circle.
Find the radius r which is the distance of O to one of the three given points.
The unknowns of equation (1) are now all known.
method 3: by determinant
The equation of the circle is given by expanding the following determinant (columns unfortunately not lined up):
x²+y² x y 1
x1²+y1² x1 y1 1
x2²+y2² x2 y2 1
x3²+y2² x3 y3 1
Using method 3, I get
(x+1)²+(y-2)²=5²
To find the equation of a circle that passes through three given points, you need to set up a system of equations using the general form of a circle equation. The general form of the circle equation is:
(x - h)^2 + (y - k)^2 = r^2
Where (h, k) represents the coordinates of the center of the circle, and r represents the radius.
Now, let's substitute the coordinates of the three given points into the general equation and obtain a system of equations.
For the point (2, 6):
(2 - h)^2 + (6 - k)^2 = r^2
For the point (-4, 6):
(-4 - h)^2 + (6 - k)^2 = r^2
For the point (-4, -2):
(-4 - h)^2 + (-2 - k)^2 = r^2
Now expand the squares and simplify:
(4 - 4h + h^2) + (36 - 12k + k^2) = r^2
(16 + 8h + h^2) + (36 - 12k + k^2) = r^2
(16 + 8h + h^2) + (4 + 8k + k^2) = r^2
Combining like terms:
2h^2 + 8h + 2k^2 - 12k + 52 = 0
Now, we can rewrite this equation in standard form:
2(h^2 + 4h + k^2 - 6k) = -52
Simplify further:
h^2 + 4h + k^2 - 6k = -26
Complete the square for both h and k:
(h^2 + 4h + 4) - 4 + (k^2 - 6k + 9) - 9 = -26
(h + 2)^2 + (k - 3)^2 = 25
Now, we see that the standard form equation of the circle is:
(x + 2)^2 + (y - 3)^2 = 25
Therefore, the center of the circle is (-2, 3), and the radius is √25 = 5.
To recap, to find the standard form equation for the circle, we substituted the given points into the general form equation, expanded the squares, combined like terms, and completed the square for both variables (h and k). Finally, we identified the center and radius from the standard form equation.