Respuesta :
Answer:
General form: [tex]4x^2+4y^2-12x-52y+72=0[/tex]
Standard form: [tex](x-\frac{3}{2})^2+(y-\frac{13}{2})^2=\frac{53}{2}[/tex]
Step-by-step explanation:
The standard form for a circle is [tex](x-h)^2+(y-k)^2=r^2[/tex] where [tex](h,k)[/tex] is the center and [tex]r[/tex] is the radius.
Our goal is to write our equation in general form: [tex]ax^2+ay^2+cx+dy+e=0[/tex].
So I'm going to use standard form right now to try and see if it helps.
Plug in the point [tex](-1,2)[/tex]:
[tex](-1-h)^2+(2-k)^2=r^2[/tex] Equation 1
Plug in the point [tex](4,2)[/tex]:
[tex](4-h)^2+(2-k)^2=r^2[/tex] Equation 2
Plug int he point [tex](-3,4)[/tex]:
[tex](-3-h)^2+(4-k)^2=r^2[/tex] 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.
[tex](-1-h)^2-(4-h)^2=0[/tex]
Add [tex](4-h)^2[/tex] on both sides:
[tex](-1-h)^2=(4-h)^2[/tex]
The implies:
[tex](-1-h)=\pm(4-h)[/tex]
This us gives us two equation to solve for [tex]h[/tex]:
[tex]-1-h=4-h[/tex] or [tex]-1-h=-4+h[/tex]
The first equation says -1=4 which is never true so we will solve the second one.
[tex]-1-h=-4+h[/tex]
Add [tex]h[/tex] on both sides:
[tex]-1=-4+2h[/tex]
Add 4 on both sides:
[tex]3=2h[/tex]
Divide 2 on both sides:
[tex]\frac{3}{2}=h[/tex]
So let's look at Equation 2 and Equation 3 with [tex]h=\frac{3}{2}[/tex] applied to them:
[tex](4-\frac{3}{2})^2+(2-k)^2=r^2[/tex]
[tex](-3-\frac{3}{2})^2+(4-k)^2=r^2[/tex]
Let's simplify them a bit by performing the addition/subtraction in the ( ):
[tex](\frac{5}{2})^2+(2-k)^2=r^2[/tex]
[tex](-\frac{9}{2})^2+(4-k)^2=r^2[/tex]
Now a little more by applying the square:
[tex]\frac{25}{4}+(2-k)^2=r^2[/tex]
[tex]\frac{81}{4}+(4-k)^2=r^2[/tex]
I will subtract these two equations now because I see it will give an equation just in terms of [tex]k[/tex] to solve:
[tex]\frac{-56}{4}+(2-k)^2-(4-k)^2=0[/tex]
Expand the binomial squares using the identity [tex](u-v)^2=u^2-2uv+v^2[/tex]:
[tex]\frac{-56}{4}+4-4k+k^2-(16-8k+k^2)=0[/tex]
Distribute:
[tex]\frac{-56}{4}+4-4k+k^2-16+8k-k^2=0[/tex]
Combine like terms:
[tex]k^2-k^2-4k+8k+\frac{-56}{4}+4-16=0[/tex]
Simplify:
[tex]4k+-26=0[/tex]
Add 26 on both sides:
[tex]4k=26[/tex]
Divide both sides by 4:
[tex]k=\frac{26}{4}[/tex]
Reduce:
[tex]k=\frac{13}{2}[/tex]
So we now have the center of the circle [tex](h,k)=(\frac{3}{2},\frac{13}{2})[/tex]. 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 [tex](\frac{3}{2},\frac{13}{2})[/tex] and [tex](4,2)[/tex].
You may use Distance Formula or Pythagorean Theorem.
Find the horizontal distance of the triangle: [tex]4-\frac{3}{2}=\frac{5}{2}[/tex].
Find the vertical distance of the triangle: [tex]\frac{13}{2}-2[/tex][tex]=\frac{9}{2}[/tex].
Now we will find the hypotenuse,[tex]c[/tex], using [tex]c^2=(\frac{5}{2})^2+(\frac{9}{2})^2[/tex].
Simplify the squares:
[tex]c^2=\frac{25}{4}+\frac{81}{4}[/tex]
Simplify by adding:
[tex]c^2=\frac{106}{4}[/tex] (This is [tex]r^2[/tex]; we don't need to find [tex]r[/tex], but I will.)
-Unnecessary for the problem; finding the radius, [tex]r[/tex]-
Take the square root of both sides:
[tex]c=\sqrt{\frac{106}{4}}[/tex]
Simplify the square root:
[tex]c=\frac{\sqrt{106}}{\sqrt{4}}[/tex]
[tex]c=\frac{\sqrt{106}}{2}[/tex]
The equation in standard form is:
[tex](x-\frac{3}{2})^2+(y-\frac{13}{2})^2=\frac{106}{4}[/tex]
(or if you simplify the fraction on the right: [tex](x-\frac{3}{2})^2+(y-\frac{13}{2})^2=\frac{53}{2}[/tex].)
Now we wanted this in general form so we will need to expand the binomial squares:
[tex]x^2-3x+\frac{9}{4}+y^2-13y+\frac{169}{4}=\frac{106}{4}[/tex]
Multiply both sides by 4 to get rid of the fractions:
[tex]4x^2-12x+9+4y^2-52y+169=106[/tex]
Reorder to put in order using commutative property:
[tex]4x^2+4y^2-12x-52y+169+9=106[/tex]
Simplify the addition on 169 and 9:
[tex]4x^2+4y^2-12x-52y+178=106[/tex]
Subtract 106 on both sides:
[tex]4x^2+4y^2-12x-52y+72=0[/tex]
The general form is [tex]4x^2+4y^2-12x-52y+72=0[/tex] .
