Question

Write the general equation for the circle that passes through the points:

Answer 1

Answer:

General form: $4x^2+4y^2-12x-52y+72=0$

Standard form: $(x-\frac{3}{2})^2+(y-\frac{13}{2})^2=\frac{53}{2}$

Step-by-step explanation:

The standard form for a circle is $(x-h)^2+(y-k)^2=r^2$ where $(h,k)$ is the center and $r$ is the radius.

Our goal is to write our equation in general form: $ax^2+ay^2+cx+dy+e=0$.

So I'm going to use standard form right now to try and see if it helps.

Plug in the point $(-1,2)$:

$(-1-h)^2+(2-k)^2=r^2$            Equation 1

Plug in the point $(4,2)$:

$(4-h)^2+(2-k)^2=r^2$            Equation 2

Plug int he point $(-3,4)$:

$(-3-h)^2+(4-k)^2=r^2$           Equation 3

I notice Equation 1 and Equation 2 will have a lot of stuff to cancel if I chose to do Equation 1 minus Equation 2. So let's do that.

$(-1-h)^2-(4-h)^2=0$

Add $(4-h)^2$ on both sides:

$(-1-h)^2=(4-h)^2$

The implies:

$(-1-h)=\pm(4-h)$

This us gives us two equation to solve for $h$:

$-1-h=4-h$ or $-1-h=-4+h$

The first equation says -1=4 which is never true so we will solve the second one.

$-1-h=-4+h$

Add $h$ on both sides:

$-1=-4+2h$

Add 4 on both sides:

$3=2h$

Divide 2 on both sides:

$\frac{3}{2}=h$

So let's look at Equation 2 and Equation 3 with $h=\frac{3}{2}$ applied to them:

$(4-\frac{3}{2})^2+(2-k)^2=r^2$

$(-3-\frac{3}{2})^2+(4-k)^2=r^2$

Let's simplify them a bit by performing the addition/subtraction in the ( ):

$(\frac{5}{2})^2+(2-k)^2=r^2$

$(-\frac{9}{2})^2+(4-k)^2=r^2$

Now a little more by applying the square:

$\frac{25}{4}+(2-k)^2=r^2$

$\frac{81}{4}+(4-k)^2=r^2$

I will subtract these two equations now because I see it will give an equation just in terms of $k$ to solve:

$\frac{-56}{4}+(2-k)^2-(4-k)^2=0$

Expand the binomial squares using the identity $(u-v)^2=u^2-2uv+v^2$:

$\frac{-56}{4}+4-4k+k^2-(16-8k+k^2)=0$

Distribute:

$\frac{-56}{4}+4-4k+k^2-16+8k-k^2=0$

Combine like terms:

$k^2-k^2-4k+8k+\frac{-56}{4}+4-16=0$

Simplify:

$4k+-26=0$

Add 26 on both sides:

$4k=26$

Divide both sides by 4:

$k=\frac{26}{4}$

Reduce:

$k=\frac{13}{2}$

So we now have the center of the circle $(h,k)=(\frac{3}{2},\frac{13}{2})$. We have multiple points to choose from so that we can find the radius. (We will find the radius by finding the distance from the center of the circle to a point on the circle.)

Let's find the the distance from $(\frac{3}{2},\frac{13}{2})$ and $(4,2)$.

You may use Distance Formula or Pythagorean Theorem.

Find the horizontal distance of the triangle: $4-\frac{3}{2}=\frac{5}{2}$.

Find the vertical distance of the triangle: $\frac{13}{2}-2$$=\frac{9}{2}$.

Now we will find the hypotenuse,$c$, using $c^2=(\frac{5}{2})^2+(\frac{9}{2})^2$.

Simplify the squares:

$c^2=\frac{25}{4}+\frac{81}{4}$

Simplify by adding:

$c^2=\frac{106}{4}$ (This is $r^2$; we don't need to find $r$, but I will.)

-Unnecessary for the problem; finding the radius, $r$-

Take the square root of both sides:

$c=\sqrt{\frac{106}{4}}$

Simplify the square root:

$c=\frac{\sqrt{106}}{\sqrt{4}}$

$c=\frac{\sqrt{106}}{2}$

The equation in standard form is:

$(x-\frac{3}{2})^2+(y-\frac{13}{2})^2=\frac{106}{4}$

(or if you simplify the fraction on the right: $(x-\frac{3}{2})^2+(y-\frac{13}{2})^2=\frac{53}{2}$.)

Now we wanted this in general form so we will need to expand the binomial squares:

$x^2-3x+\frac{9}{4}+y^2-13y+\frac{169}{4}=\frac{106}{4}$

Multiply both sides by 4 to get rid of the fractions:

$4x^2-12x+9+4y^2-52y+169=106$

Reorder to put in order using commutative property:

$4x^2+4y^2-12x-52y+169+9=106$

Simplify the addition on 169 and 9:

$4x^2+4y^2-12x-52y+178=106$

Subtract 106 on both sides:

$4x^2+4y^2-12x-52y+72=0$

The general form is $4x^2+4y^2-12x-52y+72=0$ .