Geometry question about writing equations for a circle?

Equation of circle: (x - h)² + (y - k)² = r²

Using point L(0,0):

(0-h)² + (0-k)² = r²

h² + k² = r²

Using point J(-6,0) and r²=h²+k²

(-6-h)² + (0-k)² = r²

36 + 12h + h² + k² = h² + k²

36 + 12h = 0

h = -3

Using point K(-3,3), h=-3, and r²=h²+k²=9+k²

(x - h)² + (y - k)² = r²

(-3+3)² + (3 - k)² = 9 + k²

0 + 9 - 6k + k² = 9 + k²

-6k = 0

r² = h² + k² = 9 + 0 = 9

Equation of circle, with h = -3, k = 0, r² = 9

(x + 3)² + y² = 9

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Note: if you expand, then you get:

x² + 6x + 9 + y² = 9

x² + y² + 6x = 0

1. Recommend that you plot these three points/coordinates: J(-6,0), K(-3,3), and L(0,0), on the X-Y axis, and connect these points.

2. Once you plot and connect these three points, what you have is a semi-circle, with the center of the circle located at (-3,0) and the radius of the circle is 3.

3. Standard equation of the circle: (x-h)^2 + (y-k)^2 = r^2, where r is the radius, and (h,k) is the center of the circle.

4. Using the equation in Step 3: (x-(-3))^2 + (y-0)^2 = (3)^2

5. Thus, x^2 + 3x + 3x + 9 + y^2 = 9

6. Final equation: x^2 + y^2 + 6x = 0

Hope this helps!

One approach is geometric. If you connect two of the points, J and K, or K and L, or J and L, you'll get a secant line to the circle. The perpendicular bisectors of every secant line meet at the center of the circle. So, you can use two different secant lines to find the center.

Consider the line segment from J to L. It is along the x-axis. Its midpoint is (-3,0). The line perpendicular to this line segment is vertical. The vertical line through (-3,0) is the line x = -3.

Consider the line segment from K to L. It has slope 3/-3 = -1. Its midpoint is (-3/2,3/2). The line through (-3/2,3/2) that is perpendicular to this line (its slope will be 1) is

y-3/2 = (x+3/2) ==> y = x +3.

The intersection of the two lines x=-3 and y = x+3 is the center of the circle. This point is found by substitution: x = -3, y = (-3) + 3 = 0. So the center of the circle is (-3,0).

The radius is the distance from the center to any one of the given points. Using L again (because its easy to work with), we see that the radius is 3.

The circle with raduis 3 centered at (-3,0) has equation

(x+3)^2 + y^2 = 3^2.

Note that this problem works out particularly nicely. But this approach will always work. Find two secants. Find their perpendicular bisectors. Find the point of intersection to get the center. Find the distance from the center to any given point on the circle. Write the equation.

The equation of a circle is (x-xc)^2 + (y-yc)^2 = r^2 you be attentive to that (xc = a million and yc = 3 so which you have (x-a million)^2 + (y-3)^2 = r^2 to locate t^2 plug interior the element which you be attentive to is on the circle (-2-a million)^2 + (-a million-3)^2 = r^2 (-3)^2 + (-4)^2 = r^2 9 + sixteen = r^2 25 = r^2 so the radius of the circle is 5 the terrific equation is (x-a million)^2 + (y-3)^2 = 25

By distance formula,

JK^2 = 18

KL^2 = 18 and

LJ^2 = 36

=> JK^2 + KL^2 = LJ^2

=> triangle JKL is a right triangle with LJ as hypotenuse

The cicle will have LJ as its diameter

The equation of the circle having extremeties of the diameter as

L (0, 0) and J (- 6, 0) is

(x - 0) (x + 6) + (y - 0) ( y - 0) = 0

=> x^2 + y^2 + 6x = 0

Note that all the given three points lie on this circle.

[Note: I used the formula that the equation of the circle, the endpoints of whose diameter are

(x1, y1) and (x2, y2) is (x - x1) (x - x2) + (y - y1) (y - y2) = 0.]

General equation of a circle is x²+y²+2gx+2fy+c=0 with the centre at (−g, −f).

Circle contains the points

J(-6,0) , K(-3,3) , AND L(0,0), we have

For point J(-6,0)

(−6)²+0²+2g×(−6)+2f×0+c=0 i.e 36+0²−12g+c=0

or −12g+c=−36................(i)

For point K(-3,3)

(−3)²+3²+2g×(−3)+2f×3+c=0 i.e 9+9−6g+6f+c=0

or −6g+6f+c=−18.............(ii)

For point L(0,0)

(0)²+0²+2g×(0)+2f×0+c=0 i.e c=0 ..........(iii)

Substituting (iii) in (i), we get

−12g+0=−36 →g=3.........(iv)

From (ii), (iii) and (iv)

−6×3+6f+0=−18→ f =0

Hence the equation of the required circle is

x²+y²+2×3x+2×0×y+0=0

or x²+y²+6x=0

Generalised Circle has equation

x^2+y^2+gx+hy+k=0

Insert J

36-6g+k=0

Insert L

k=0

Insert K

9+9-3g+3h=0

So g=6

3h=0

So h=0

x^2+y^2+6x=0

It's mostly about organizing your Xs and Ys. When you get the 3 points where they belong, then it's just about seeing a full circle and how the points you have fit/complete it.