use the information provided to write the general conic form equation of each circle.

1. three points on the circle, ( 7,-4),(-3,-4), and (6,-7)

please help, I do not understand how to dod this problem.

xxJenny1213xx Mar 23, 2020

#1**+2 **

Hey Jenny! See you got some conics to deal with!

Let's start with the general equation of a circle, the one that most people are taught:

(x - a)^{2} + (y-b)^{2} = r^{2}, where (a,b) is the center of the circle, and r is the radius of the given circle.

Because the question gives us 3 points, we know they all must satisfy this equation.

Let's pick the pair of points (7,-4) and (-3,-4) to plug in(there's a good reason for picking this pair. Do you see why?)

We can first look at (7,-4). This gives us:

(7-a)^{2} + (-4-b)^{2} = r^{2}

Let's then look at (-3,-4). It'll start to become clear why I picked these two points. We get:

(-3-a)^{2} + (-4-b)^{2} = r^{2}

We realize that we can then set these two equations equal to each other, to get:

(7-a)^{2} + (-4-b)^{2} = (-3-a)^{2} + (-4-b)^{2}

Now here's where our selection of points comes in handy. Realize we can just subtract

(-4-b)^{2} on both sides, which gives us only a, allowing us to solve for a without getting a nasty(ier) expression than if we chose another pair.

We get:

(7-a)^{2} = (-3-a)^{2}

Expanding, we get:

49-14a + a^{2} = 9+6a + a^2

Combining like terms and subtracting appropriately, we get:

40 = 20a

**a = 2**

To get b, let's create another statement with two different points, say (7, -4) and (6,-7)

We get:

(7-2)^{2} + (-4 -b)^{2} = (6-2)^{2} + (-7-b)^{2}

This gives us:

25 + (-4 -b)^{2} = 16 + (-7-b)^{2}

9 + 16 + 8b + b^{2} = 16 + 49 + 14b + b^{2}

Combining like terms and subtracting where is necessary, we obtain the equation:

6b = -40

**b = -20/3**

**Our circle's equation is then:**

**(x - 2) ^{2} + (y+20/3)^{2} = r^{2}**

I'll leave finding the radius to you(it should be fairly simple, just plug in one of the points to find it, and then take the square root of the result.

jfan17 Mar 23, 2020