**Implicit function symbolically and graph: **if x and y are **so mixed**, cannot be expressed in terms of the **independent variable** is called an** implicit function **of x.

** For example:**

**(i) x²+x.y+y²=2**

**(ii) x.y² -y+9 /x.y =1**

are** implicit function** of **x and y.**

**Symbolically,** it has written as** f(x, y)=0**

**Graphs of Implicit function:**

**(a)** graph of **circle** of the form** x²+y²=a²**

**Example 1: implicit function symbolically and graph**

**Graph of circle x²+y²=4 (1)**

**Solution:**

The graph of equation **x²+y²=4** is a **circle of radius 2.** Centered at **origin** and hence there are **vertical lines** that** cut** the graph more than once, this can also been seen **algebraically** by solving (1) for** y in terms of x.**

**y = ±√4-x²**

The equation does not define y as a function of x.

**For example**:

**if x=1 then y= ±√3**

hence **(1,+√3) and (1,-√3)** are two point on the** circle** and a vertical line passes through these** two points.**

We can regard the **circles as the union** of two **semé circles.**

**y =√4-x² and y=– √4-x²**

Each of which defines** y as a function of x**.

We observe that if we replace** (x, y)** in turn by** (-x, y) and (-x, y)and (-x, -y),** there is no change in the given equation. Hence, the graph is** symmetric** with respect to the **y-axis and x-axis and the origin.**

**x=0 implies y²=4 ⇒ y=±2**

**x=1 implies y²=3 ⇒ y=±√3**

**x=2 implies y²=0 ⇒ y=0**

by assigning values of x. we fined the value of y. so we prepare a table for some value of x and y satisfying equation (1)

**Radius 2**

**if when x=2 then y=0 **

**if when y=2 then x=0**

x |
0 |
1 |
√3 |
2 |
-1 |
-√3 |
-2 |

y |
±2 |
±√3 |
±1 |
0 |
±√3 |
±1 |
0 |

**plotting the graph (x, y)** and connecting them with the **smooth curve** as shown in figure. We get a **graph of the figure.**

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