Answer:
The equation of circle becomes \(\left(x-3\right)^2+\left(y-\left(-7\right)\right)^2=\left(10\sqrt{2}\right)^2\)
Step-by-step explanation:
Given equation of circle \(x^2+y^2-6x+14y=142\)
We have to write the equation of this circle written in standard form.
The standard equation of a circle with (h,k) as center and radius r, is given by
\((x-h)^2+(y-k)^2=r^2\\\)
For the given equation of circle \(x^2+y^2-6x+14y=142\)
making perfect squares ,
Rearranging terms, we have,
\(x^2+y^2-6x+14y=142\) as \(x^2-6x+y^2+14y=142\)
Applying \((a+b)^2=a^2+b^2+2ab\\ (a-b)^2=a^2+b^2-2ab\)
We have \(x^2-6x+y^2+14y=142\)
To make \(x^2-6x\) a perfect square add and subtract 3²= 9 , we get,
\(x^2-6x+9-9+y^2+14y=142\)
Similarly, To make \(y^2+14y\) a perfect square add and subtract 7²= 49 we get,
\(x^2-6x+9-9+y^2+14y+49-49=142\)
Simplify, we get,
\((x-3)^2+(y+7)^2=142+49+9\)
Simplify , we have,
\((x-3)^2+(y+7)^2=200\)
Thus, the equation of circle becomes \(\left(x-3\right)^2+\left(y-\left(-7\right)\right)^2=\left(10\sqrt{2}\right)^2\)
