The general form of the equation of a circle is x2 + y2 + 42x + 38y − 47 = 0. The equation of this circle in standard form is .
The center of the circle is at the point , and its radius is units.
The general form of the equation of a circle that has the same radius as the above circle is .
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To transform general form to a standard form, you can use completing the square method or you can convert this equation x² + y² + Dx + Ey + F = 0 to x² + y² + (-2h)x + (-2k)y + (h²+k²-r²) = 0

Given: x2 + y2 + 42x + 38y − 47 = 0

Required: Find the values of h, k, and r using this formula x² + y² + (-2h)x + (-2k)y + (h²+k²-r²) = 0

Solution:

-2h = 42

  h = 21

-2k = 38

  k = 19

-47 = h^2 + k^2 - r^2

-47 = 21^2 + 19^2 - r^2

r^2 = 849

   r = √849

Answers:

1.) Standard form : (x+21)² + (y+19)² = 849

2.) Center of the circle : (h,k) = (21,19)

3.) Radius of the circle : r = √849

4.) Equation of a circle that has the same radius as the above circle is : x^2 + y^2 - 50x - 30y + 1 = 0

You can solve the last one through trial and error. Its h - 25; k - 15.

1 = h^2 + k^2 - r^2

1 = -25^2 + -15^2 - r^2

r^2 = 849

   r = √849  :  which has the same radius as the given equation.

b, i believe, but not sure.

answer: yes, since the line has a steeper rate of change than the exponential the y-values will be greater and they will only intersect one time.