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Find the center and radius of the circle with the given equation:x² + y² - 8x + 10y = -16

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ANSWER


\begin{gathered} Center:(4,-5) \\ Radius:5 \end{gathered}

Step-by-step explanation

We want to find the center and radius of the circle with the given equation:


x^2+y^2-8x+10y=-16

To do this, we have to write the equation in standard form:


(x-h)^2+(x-k)^2=r^2

where (h, k) = center

r = radius

To do this, complete the square for both x and y variables of the equation.

Let us do that now:


\begin{gathered} x^2-8x+y^2+10y=-16 \\ x^2-8x+(-(8)/(2))^2+y^2+10y+((10)/(2))^2=-16+(-(8)/(2))^2+((10)/(2))^2 \\ x^2-8x+16+y^2+10y+25=-16+16+25 \\ x^2-8x+16+y^2+10y+25=25 \end{gathered}

Finally, factorize both the x and y variables:


\begin{gathered} x^2-4x-4x+16+y^2+5y+5y+25=25 \\ x(x-4)-4(x-4)+y(y+5)+5(y+5)=25 \\ (x-4)(x-4)+(y+5)(y+5)=25 \\ (x-4)^2+(y+5)^2=5^2 \end{gathered}

Therefore, the center and radius of the circle are:


\begin{gathered} Center:(4,-5) \\ Radius:5 \end{gathered}

That is the answer.

User MathCatsAnd
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