Question

1. find the general equation of $(x - 2)^2+(y - 5)^2 = 20$.
2. what is the general equation of a circle with center at $(-4,7)$ and passing through $(0,4)$?

Explanation:

Step1: Expand the first - circle equation

Expand \((x - 2)^2+(y - 5)^2=20\).
Using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), we have \((x - 2)^2=x^{2}-4x + 4\) and \((y - 5)^2=y^{2}-10y + 25\).
So \(x^{2}-4x + 4+y^{2}-10y + 25=20\), which simplifies to \(x^{2}+y^{2}-4x-10y + 4 + 25-20=0\), and further to \(x^{2}+y^{2}-4x-10y + 9 = 0\).

Step2: Find the radius of the second - circle

First, find the radius \(r\) of the circle with center \((-4,7)\) passing through \((0,4)\) using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
Here, \(x_1=-4,y_1 = 7,x_2=0,y_2 = 4\). Then \(r=\sqrt{(0+4)^2+(4 - 7)^2}=\sqrt{16 + 9}=\sqrt{25}=5\).

Step3: Write the standard - form equation of the second - circle

The standard - form of a circle equation is \((x - a)^2+(y - b)^2=r^{2}\), where \((a,b)\) is the center and \(r\) is the radius.
For a center \((-4,7)\) and \(r = 5\), the equation is \((x + 4)^2+(y - 7)^2=25\).

Step4: Expand the second - circle equation

Expand \((x + 4)^2+(y - 7)^2=25\).
Using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), \((x + 4)^2=x^{2}+8x+16\) and \((y - 7)^2=y^{2}-14y + 49\).
So \(x^{2}+8x+16+y^{2}-14y + 49=25\), which simplifies to \(x^{2}+y^{2}+8x-14y+16 + 49-25=0\), and further to \(x^{2}+y^{2}+8x-14y + 40=0\).

Answer:

1. \(x^{2}+y^{2}-4x-10y + 9 = 0\)
2. \(x^{2}+y^{2}+8x-14y + 40=0\)