Question

How do you find the equation of the circle given radius = $ 4; $ and center = $ \left( { - 2,3} \right)? $

Answer 1

Hint: Circle: if we take a point as center then draw an infinite number of points around that center having equal distance from the center. This is known as a circle.
By using this property. General equation of circle is:
  $ {\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2} $ ….. $ (1) $
Where, $ \left( {h,k} \right) $ is the center $ \& \,r $ is the radius

Complete step by step solution:
As given in question,
Here center $ \left( {h,k} \right) = \left( { - 2,3} \right) $
We get, $ h = - 2\& k = 3 $ and Radius $ (r) = 4 $
As general equation of circle is $ {\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2} $
By keeping values from above in the equation. We get,
  $ {\left( {x - \left( { - 2} \right)} \right)^2} + {\left( {y - 3} \right)^2} = {(4)^2} $
  $ \Rightarrow {(x + 2)^2} + {(y - 3)^2} = 16 $
  $ \Rightarrow ({x^2} + {2^2} + 4x) + ({y^2} + {3^2} - 6y) = 16 $
  $ \Rightarrow {x^2} + 4 + 4x + {y^2} + 9 - 6y = 16 $
  $ \Rightarrow {x^2} + {y^2} + 4x - 6y + 13 - 16 = 0 $ v
  $ \Rightarrow {x^2} + {y^2} + 4x - 6y - 3 = 0 $
Hence the equation of circle for given equation will be $ {x^2} + {y^2} + 4x - 6y - 3 = 0 $
So, the correct answer is “ $ {x^2} + {y^2} + 4x - 6y - 3 = 0 $ ”.

Note: General equation is formed for different type entities. We can identify from the given general equation about the given object and its basic properties like length, radius, coordinate, vertex etc. It is mostly used in two-dimensional geometry.