Question

Find the equation of circle whose center is $( - 3,1)$ and which pass through the point $(5,2)$

Answer 1

Hint: Coordinate geometry is defined as the study of the geometry using the coordinates points. Using coordinate geometry we find the distance between the two points. Circle in the coordinate geometry, the equation of circle is given by
$\therefore {(x - {x_0})^2} + {(y - {y_0})^2} = {r^2}$.
If the coordinate satisfies the equation then the point is in the circle.
Here,
P=Perimeter of circle
R=radius of circle.

Complete step-by-step solution:
Given,
Center of circle$ = ( - 3,1)$
Pass through$ = (5,2)$
Radius of circle=?
As we know that radius is,
$\therefore r = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
Put the value
$ \Rightarrow r = \sqrt {{{\{ 5 - ( - 3)\} }^2} + {{(2 - 1)}^2}} $
Simplify
$ \Rightarrow r = \sqrt {{8^2} + {1^2}} $
$ \Rightarrow r = \sqrt {64 + 1} $
$ \Rightarrow r = \sqrt {65} $
Radius of the circle is $\sqrt {65} $
Cartesian equation of circle is given by
$\therefore {(x - {x_0})^2} + {(y - {y_0})^2} = {r^2}$
Put the value.
$ \Rightarrow {(x - ( - 3))^2} + {(y - 1)^2} = {(\sqrt {65} )^2}$
$ \Rightarrow {(x + 3)^2} + {(y - 1)^2} = 65$
$ \Rightarrow {x^2} + {y^2} + 6x - 2y + 10 - 65 = 0$
$ \Rightarrow {x^2} + {y^2} + 6x - 2y - 55 = 0$
The required equation for circle is ${x^2} + {y^2} + 6x - 2y - 55 = 0$

Note: Here we could also find the equation of circle by assuming the general equation of circle and then we put the given points and apply the algebraic operation for finding the unknown variables but that would be a lengthy approach.