Question

give the center and radius of the circle represented by the equation. 49x^2 + 49y^2 - 56x + 56y - 32 = 0 the center of the circle is . (type an ordered pair, using integers or fractions.) the radius of the circle is . (type an integer or a simplified fraction.)

Explanation:

Step1: Divide the equation by 49

Divide the entire equation $49x^{2}+49y^{2}-56x + 56y-32 = 0$ by 49 to get $x^{2}+y^{2}-\frac{8}{7}x+\frac{8}{7}y-\frac{32}{49}=0$.

Step2: Complete the square for x - terms

For the x - terms in $x^{2}-\frac{8}{7}x$, we have $(x - \frac{4}{7})^{2}=x^{2}-\frac{8}{7}x+\frac{16}{49}$.

Step3: Complete the square for y - terms

For the y - terms in $y^{2}+\frac{8}{7}y$, we have $(y+\frac{4}{7})^{2}=y^{2}+\frac{8}{7}y+\frac{16}{49}$.

Step4: Rewrite the equation

Rewrite the equation $x^{2}+y^{2}-\frac{8}{7}x+\frac{8}{7}y-\frac{32}{49}=0$ as $(x - \frac{4}{7})^{2}- \frac{16}{49}+(y+\frac{4}{7})^{2}-\frac{16}{49}-\frac{32}{49}=0$.

Step5: Simplify the equation

Simplify the above - equation: $(x - \frac{4}{7})^{2}+(y+\frac{4}{7})^{2}=\frac{16 + 16+32}{49}=\frac{64}{49}$.

Answer:

The center of the circle is $(\frac{4}{7},-\frac{4}{7})$.
The radius of the circle is $\frac{8}{7}$.