Question

How do you find the equation of a circle in standard form given ${x^2} + {y^2} - 10x + 6y + 30 = 0$?

Answer 1

Here the question needs to be solved for the standard equation of the circle, and here the standard form of equation is written after expansion, in the standard form we can easily get the centre coordinates and the radius of the circle. Here we have to do the conversion from the given question to the standard form.

Formulae Used:
${(x - a)^2} + {(y - b)^2} = {r^2}$It is the standard equation for a circle of radius “(a,b)” and radius is equal to “r”.

Complete step by step solution:
Here the given question is ${x^2} + {y^2} - 10x + 6y + 30 = 0$
Now first we have to see the standard equation of circle for any centre “(a,b)” and radius “r”, on solving we get:
$\Rightarrow {(x - a)^2} + {(y - b)^2} = {r^2}$
Now we have to convert the given question into this form, above written expression is just the simplified form the standard form, so we can simplify our equation by making perfect square for both the variables, and the rest remaining numerical value will be our radius for the circle, on solving we get:

$$\Rightarrow {x^2} + {y^2} - 10x + 6y + 30 = 0 \\\ \Rightarrow {x^2} + {y^2} - 2(5)x + 2(3)y + 25 - 25 + 9 - 9 + 30 = 0 \\\ \Rightarrow {x^2} - 2(5)x + 25 + {y^2} + 2(3)y + 9 - 25 - 9 + 30 = 0 \\\ \Rightarrow {(x - 5)^2} + {(y + 3)^2} = 4 = {2^2} \\\ \Rightarrow {(x - 5)^2} + {(y + 3)^2} = {2^2} \\\$$

Here we finally get the standard equation for the given circle, whose radius is “(5,-3)” and radius of the circle is two.

Note: The given question needs to be solved for converting the expression from expanded form to the standard form, for which we have used the perfect square making method, but here we can also directly expand the standard equation and on comparison we can directly write the standard equation of the given circle.