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