The standard equation of a circle with radius r is the relation between the center of the circle, (h,k), and any point that lies on the circle, (x,y).
The distance between the center and any point on the circle is always the radius, r.
The equation of a circle can be derived from the Pythagorean Theorem.
Consider a circle centered at (h,k) such that the point (x,y) lies on the circle.
A right triangle can be created by drawing a horizontal segment to the right of (h,k) and a vertical segment down from (x,y).
The hypotenuse of the triangle has the length r. To find the length of the legs, the coordinates of the third vertex are marked.
The lengths of the legs can now be expressed as the difference between the x- and y-coordinates of the vertices.
Therefore, the length of the legs can be expressed as (x−a) and (y−b). Since the triangle is a right triangle, the Pythagorean Theorem applies. a2+b2=c2 Here, a and b are the lengths of the legs and c is the hypotenuse. Substituting the stated values for the triangle drawn above yields the standard equation of a circle. (x−h)2+(y−k)2=r2
Graph the circle from the equation (x+3)2+(y−1)2=22.
If a circle's center and radius are known the circle can be drawn in a coordinate plane by writing the given equation in standard form. (x−h)2+(y−k)2=r2 From the given equation, it can be seen that the center of the circle is (-3,1) and its radius is 2. First, the center can be marked.
The circle can now be drawn using a compass. Set the compass to the length of 2 units by placing the needle at the center C and the pencil at any point 2 units away from C.
Now, draw the corresponding circle with the pencil on the compass.
The following graphs shows the circle with the given equation.
The standard equation for a circle can be written using the radius, the center, and a point on the circle.
The equation of the circle shown above can be written using the following method.
The general form of the equation of a circle is (x−h)2+(y−k)2=r2, where (h,k) is the center of the circle and r is the radius. Of the values needed to write the equation, the center is given: P(-2,1). Thus, h=-2 and k=1. The radius of the circle remains to be found.
The radius of the circle can be drawn from P to Q.
Given a circle with center at (-1,1) and the radius 5 units. Determine if the point (-3,3) lies on the circle.
The standard equation of a circle consist of two factored perfect square trinomials on the left-hand side. (x−h)2and(y−k)2 By completing the square on the x- and y-terms, perfect square trinomials can be created. The terms on the left-hand side of the equation can be rearranged so that the terms with the same variables are next to each other. x2+y2+10x−4y=20⇔(x2+10x)+(y2−4y)=20 It is necessary to complete the square on (x2+10x) and (y2−4y).
The constants should now be added to the equation. (x2+10x+25)+(y2−4y+4)=20+25+4 The terms on the right-hand side can be added. (x2+10x+25)+(y2−4y+4)=49