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Question: The general form of unit circle is: A.\({x^2} + {y^2} = {r^2}\) B.\({x^2} - {y^2} = {r^2}\) C....

The general form of unit circle is:
A.x2+y2=r2{x^2} + {y^2} = {r^2}
B.x2−y2=r2{x^2} - {y^2} = {r^2}
C.x2+y2=1{x^2} + {y^2} = 1
D.x2+y2=2{x^2} + {y^2} = 2

Explanation

Solution

Hint: Write the general form of the equation of the circle with radius (h,k)\left( {h,k} \right) and centre rr is
(x−h)2+(y−k)2=r2{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}. To find the general form of the unit circle, take the centre as (0,0)\left( {0,0} \right) and radius as 1 unit. Simplify the equation to write the general form of the unit circle.

Complete step-by-step answer:
A circle is a locus of points that has a fixed distance from a fixed point.
The general form of a circle of centre (h,k)\left( {h,k} \right) with radius rr is (x−h)2+(y−k)2=r2{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}
We have to find the general equation for the unit circle, that is the circle with radius as 1 unit.
Let the centre be at origin and radius as 1 unit.
On substituting the value of (h,k)\left( {h,k} \right) as (0,0)\left( {0,0} \right) and radius as 1, we get
(x−0)2+(y−0)2=12{\left( {x - 0} \right)^2} + {\left( {y - 0} \right)^2} = {1^2}
On simplifying the equation we get,
x2+y2=12{x^2} + {y^2} = {1^2}
Hence the general form of the unit circle is x2+y2=1{x^2} + {y^2} = {1}
Hence, option C is the correct answer.

Note: Since, a circle is a locus of points that has a fixed distance from a fixed point , the equation can be derived using distance formula from a fixed point known as centre and fixed distance is known as radius of the circle. The general form of the circle is also written as, x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2gx + 2fy + c = 0, where coordinates of the centre is (−g,−f)\left( { - g, - f} \right) and the radius is g2+f2−c\sqrt {{g^2} + {f^2} - c} .