Learn the general equation of a circle when the center is at origin and when it is not in origin. Circle equation can be derived using Pythagoras theorem as well. Solve questions at BYJU’S.
Let's see some examples based on the standard form. ADVERTISEMENTExample 1: Find out the radius and center of a circle from the given equation. Also, draw the graph. (x-2)2+(y-3)2=4 Solution: The given equation is, (x-2)2+(y-3)2=4. ...
EquationofCircles GeneralForm 222 ()()xaybr whereristheradiusand(a,b)isthecentre. Example Acircleofcentre(5,−4)andradius3has equation 222 22 (5)((4))3 (5)(4)9 xy xy Centre(−1,2),radius7 222 22 ((1))(2)7 (1)(2)49 xy xy Centre(0,0),radius1 22 1xy Centre(4,...
Radius –Radius is the distance from the centre of a circle to any point on the boundary of the circle. Diameter –Diameter is the line from one point on the boundary of the circle to another point and passing through the centre of the circle. It is twice the length of the radius. Ch...
Here, the radius of the circle,r=3 units Since, the circle touches both the axes and completely lies in the third quadrant, its centre is at(-3,-3).Now, the equation of the circle is(x-h)²+(y-k)²=r²or,{x-(-3)}²+{y-(-3)}²=3²or,(x+3)²+(y+3)²...
Equation of a Unit CircleThe general equation of a circle is of the form (x−a)2+(y−b)2=r2, where the center of the circle is (a, b) and the radius is r. A unit circle in the x-y plane is formed with a center at origin (0,0) and radius 1. Thus...
Equation of a cirle. How to express the standard form equation of a circle of a given radius. Practice problems with worked out solutions, pictures and illustrations.
Example 3: Find the equation of a circle with centre at (4, -4) that touches both the axes. 相关知识点: 试题来源: 解析 Here, the centre of the given circle is at(4,-4). As the circle touches both the axes, its radius is 4 units. The equation of the circle is(x-h)²+(y...
ExampleQuestion:1 (a,b) r (i)Findtheequationofthecirclewith centre(4,-3)andradius6. (ii)Doesthepoint(3,1)lieon,inside,or outsidethecircle? (i)Thecirclehasequation: (x–4) 2 +(y–(-3)) 2 =6 2 (x–4) 2 +(y+3) 2 =36 (ii)Substituting(3,1)intotheequation: (3–4) 2...
Consider now a new example. Let x2+y2+4x−6y−12=0. To find the center-radius form of the above circle, we are going to use a technique called completing the square. First, we are going to rewrite the equation as x2+4x+y2−6y=12. Note that if we add 4 to both sides...