Do you have difficulty in finding the center and the radius of the general form of a circle's equation? If yes, then relax. ππ

Be cool!π¨π¦π§πΉ

**Grouping of terms** with variables *x* and *y* ,and **completing the squares** are not needed using these **shortcuts**.

Are you ready to learn? βΊππβΊ

We usually look for the faster manual way or method on how to find the **center** and the **radius** of the general form of circle's equation, don't we?

The **general form** of the circle's equation is **xΒ²+yΒ²+ax+by+c=0**.π

Here is an example. π

**Sample Problem**:

What is the **center** and the **radius** of

**xΒ²+yΒ²-2x-12y-63=0** ? πβ³

Given: **a**= -2, **b**= -12, and **c**= -63

**Solution**:

*Shortcut in finding the coordinates of the circle's center*

**C(h,k)= (a/-2, b/-2)** π**FORMULA**π¨

Substitute -2 for **a** and -12 for **b**.

C(h,k)= (-2/-2),-12/-2)

**C(h,k)= (1,6)** βπ *center*

*Shortcut in finding the circle's radius length*

Let *sqrt* stands for the *square root*.

**r = sqrt(hΒ²+kΒ² -c)** π

**FORMULA**π¨

Substitute 1 for **h**, 6 for **k**, and -63 for **c**.

r = *sqrt*(1Β² +6Β² -(-63))

r = *sqrt*(1+ 36 +63)

r = *sqrt*(100)

**r = 10 units** βπ *radius length*

Hence, the circle's *center* is at point **C(1,6)** and its *radius* length **10 units**.

Now, you try! βΊ

**Exercises**: π

Solve the center and the radius length of each circle's equation.

```
1. xΒ²+yΒ²-4x+8y-9=0
2. xΒ²+yΒ²+10x-18y+11=0
3. xΒ²+yΒ²+2y-3=0
β°
```

*You can also calculate each circle's diameter, circumference and area given its computed radius length.*

Author: **John Paul L. Hablado**, *LPT*

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