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Question

Question: How do you graph \({x^2} + {y^2} = 36\)...

How do you graph x2+y2=36{x^2} + {y^2} = 36

Explanation

Solution

Here we use the standard form of a circle that is (xa)2+(yb)2=r2{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2} where (a,b)\left( {a,b} \right) are coordinates of the circle and rr is the radius of the circle. By using this standard form of a circle we will have the required result.
Formula used: The equation of a circle is (xa)2+(yb)2=r2{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}

Complete step by step answer:
We need to draw the graph of the given equation
So we are proceeding in that way
The given question can be written in the form of the standard form of a circle in the following way:
(x0)2+(y0)2=62{\left( {x - 0} \right)^2} + {\left( {y - 0} \right)^2} = {6^2}
Here (0,0)\left( {0,0} \right) is the coordinate of the centre and 66 is the radius of the circle.
Hence, the circle can be drawn in the following manner:
Now, plotting the points on the curve and joining it we get a circle as shown in the figure above.

Here the centre lies on the coordinate (0,0)\left( {0,0} \right) and the radius is of 66 units.

Note:
The given equation x2+y2=36{x^2} + {y^2} = 36 represents the locus of a point that moves in such a way that its distance from the point (0,0)\left( {0,0} \right) is always equal to66.
The question can also be solved in a way by putting the values of xx and yy equal to 00 simultaneously.
Putting x=0x\, = \,0, we will get y2=36{y^2} = 36 which implies that the value of yy is equal to +6or6 + 6\,or\, - 6
Hence, the coordinates come out to be (0,6)\left( {0,6} \right) and (0,6)\left( {0, - 6} \right)
Similarly Putting y=0y\, = \,0 we will get x2=36{x^2} = 36 which implies that the value of xx is equal to +6or6 + 6\,or\, - 6.
Hence, the coordinates come out to be (6,0)\left( {6,0} \right) and (6,0)\left( { - 6,0} \right)
General form of the circle is x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2gx + 2fy + c = 0 , where (g,f)( - g, - f) is the centre and g2+h2c\sqrt {{g^2} + {h^2} - c} is the radius of the circle .