The **standard form** of the equation of the circle is: (x – h)^{2} + (y – k)^{2} = r^{2 }where the r is the radius and the point (h, k) is the center. If the center of the circle is the origin (0, 0), then the equation becomes x^{2} + y^{2} = r^{2}. I am sure you know (x- 0)^{2} =x^{2} and (y- 0)^{2} =y^{2}

Notice we have **subtraction inside the parentheses,** but *addition between them*!

If the **center is point ( 2, -5)** then that means h = 2 and k= -5. Let **15 be the radius** of the circle. Our equation becomes

**(**x **– **(2) **)**^{2} + **(**y **–** (-5) **)**^{2} = 15^{2 }. Simplify inside the parentheses first!

#### (x – 2)^{2} + (y +5)^{2} = 225.

Now let us use algebra to square each pair of parentheses.

#### Be careful! (x – 2)^{2 }is NOT x^{2} + 4.

To square these I suggest writing them twice and then multiply. SEE: (x – 2 )(x – 2) = x^{2} -2x -2x +4 = x^{2} -4x+4. Now do the same for the (y +5)^{2 } You will have y^{2} +10y +25.

Add together: x^{2} – 4x+4 + y^{2} +10y +25 = 225. And simplify.

x^{2} -4x + y^{2} +10y +29= 225 becomes

x^{2} -4x + y^{2} +10y -196=0 . This is the **GENERAL form** of the circle.