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Question: How do you find the center and radius of the circle \({{x}^{2}}+{{y}^{2}}=4\)?...

How do you find the center and radius of the circle x2+y2=4{{x}^{2}}+{{y}^{2}}=4?

Explanation

Solution

A circle is the locus of all points drawn with a fixed length from a fixed point. This fixed length is called the radius of the circle and the fixed point is called the center of the circle. The most common and important center - radius form of the circle equation is
(x−h)2+(y−k)2=r2{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}, where the coordinate pair (h,k)\left( h,k \right) will give the coordinates of the center of the circle and rr gives the radius of the circle . This form of the equation is very important because it helps in finding the center as well as the radius of the circle. This is also a general form of the equation of a circle and it can be obtained if we are given the coordinates of the center as well as the radius of the circle.

Complete step by step solution:
The equation of the circle given to us is
x2+y2=4{{x}^{2}}+{{y}^{2}}=4
This equation can be rewritten as
(x−0)2+(y−0)2=(2)2{{(x-0)}^{2}}+{{(y-0)}^{2}}={{(2)}^{2}}
And the general equation of the circle is
(x−h)2+(y−k)2=r2{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}
Now comparing both the equations , we get

h=0\text{ }, \\\ k=0 \\\ \end{array}$$ and $$r=2$$ **Therefore the center of the given circle is $$\left( h,k \right)$$ which is equal to $$\left( 0,0 \right)$$ that is the origin . And the radius $r$ of the given circle is $2$.** **Note:** We can draw a right-angled triangle with the center of the circle as one of the points of the triangle and one point lying on the circumference such that the radius r of the circle is equal to the hypotenuse of the right-angled triangle. If the coordinates of the point on the circumference of the circle is $$\left( x,y \right)$$. Then by using Pythagoras theorem, we get ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$ And this equation satisfies any point on the circle.