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The standard form of the equation of a circle is. where (a ,b) are the coordinates of the centre and r, the radius. Substitute these values into the standard equation. ⇒**(x+2)2+(y+3)2=9** is the circle’s equation.

So, if the center is (0,0) and the radius is 6, an equation of the circle is: **(x-0) ^{2} + (y-0)^{2} = 6^{2}**.

**x2+y2=4**.

Simple. Take (positive) sqrt on both sides, then equation **says distance between all points (x,y) from a fixed point (h,k) is r**. This clearly indicates figure is a circle.

Equation for the circle with center at (2,-1) with radius 5 is, **(x-2) ^2 + (y+1)^2 = 25**.

**x2+y2=8**.

Explanation: A general circle with centre (a,b) and radius r has equation **(x−a)2+(y−b)2=r2** .

So in this case since the centre is the origin, it implies that a=b=0 , and the radius r=9⇒r2=92=81 . Thus the equation reduces to **x2+y2=81** .

The general equation for a circle, centered at (a,b) with radius r , is the set of all points (x,y) such that **(x−a)2+(y−b)2=r2 ( x − a ) 2 + ( y − b ) 2 = r 2** . Pi (π ) is the ratio of a circle’s circumference to its diameter.

The general equation of a circle is **(x – a)^2 + (y – b)^2 = r^2**, where the center is at (a , b) and the radius is r. Now we have the center of the circle at (0,0). It passes through the point (3,4).

What is the equation of the circle whose center is in the origin and has a diameter of 12? **x² + y² = 36** is the answer!

The formula for the equation of a circle is **(x – h) ^{2}+ (y – k)^{2} = r^{2}**, where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle.

(x−h)2+(y−k)2=r2 . Accordingly, the **reqd**.

⇒**(x−2)2+(y+3)2=25** is the equation of the circle.

Summary: The equation which represents a circle with a center at (-4, 9) and a diameter of 10 units is **(x + 4) ^{2} + (y – 9)^{2} = 25**.

The equation of a circle with center (h,k) and radius r units is **(x−h)2+(y−k)2=r2** .

Q. In the equation **(x+2) ^{2}+(y-3)^{2}=4**, the radius of the circle is… Which equation has a center (0, 0) and a radius of 3? Q.

In this lesson, you learned the equation of a circle that is centered somewhere other than the origin is **(x−h)2+(y−k)2=r2**, where (h,k)is the center.

**If g2+f2−c=0**, then it’s a point circle. If g2+f2−c>0, then it’s a real circle. If g2+f2−c

The equation of a circle is **a way to express the definition of a circle on the coordinate plane**. If the center of the circle is at the origin of the coordinate plane, the equation is where r is the radius. … Using the Completing the Square technique converts the equation to an easier form.

The center is at the origin. Where r is the radius and (h,k) is center. If either -h or **-k is missing**, then its value must be 0. Thus, if both are missing the circle must be centered at the origin, (0,0).

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