Question

THIS IS NOT A TEST THIS A STUDY GUIDEThe equation of this circle in standard form isA. (x - 2)^2 + (y + 3)^2 = 5B. (x - 2)^2 + (y + 3)^2 = 18C. (x - 4)^2 + (y + 6)^2 = 24D. (x + 4)^2 + (y - 6)^2 = 24E. (x + 2)^2 + (y - 3)^2 = 35The center of the circle is at the point __A. (-14,21)B. (-2,3)C. (2,-3)D. (-4,6)E. (4,-6)And it's radius is ___ units.A. 5 1/2B. 3(2 1/2)C. 2(6 1/2)D. 35 1/2

Answer 1

Answer

Explanation

Given equation:

$7x^2+7y^2-28x+42y-35=0$

Step-by-step explanation:

The equation of a circle in standard form is given by:

$\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ whereh,(h,k)\text{ }are\text{ }the\text{ }coordinates\text{ }of\text{ }the\text{ }centre,\text{ }and\text{ } \\ r\text{ }is\text{ }the\text{ }radius \end{gathered}$

Divide the give equation through by 7:

$\begin{gathered} \frac{7x^2}{7}+\frac{7y^2}{7}-\frac{28x}{7}+\frac{42y}{7}-\frac{35}{7}=\frac{0}{7} \\ \\ x^2+y^2-4x+6y-5=0 \end{gathered}$

TCombine the like terms:

$x^2-4x+y^2+6y=5$

Using perfect square on each variable:

$\begin{gathered} (x^2-4x+4)+(y^2+6y+9)=5+4+9 \\ \\ (x-2)^2+(y+3)^2=18 \\ \\ \therefore(h,k)=(2,-3) \\ \\ r^2=18 \\ \Rightarrow r=\sqrt{18}=3\sqrt{2} \end{gathered}$

Therefore, the equation of this circle is