Write the equation of the circle tangent to the coordinate axes and passing through the point (8; -4).

1. A circle centered at point O1 (4; -4) and radius R = 4 touches the coordinate axes at points (4; 0) and (0; -4) and passes through point (8; -4).

2. Since the equation of a circle with radius r and center at the point (x0; y0) is:

(x – x0) ² + (y – y0) ² = r²,

then for the drawn circle we obtain the following equation:

(x – 4) ² + (y + 4) ² = 16.

3. General solution. Find the radius of the circle:

O (r; -r) and (8; -4);

(r – 8) ² + (-r + 4) ² = r²;

r² – 16r + 64 + r² – 8r + 16 = r²;

r² – 24r + 80 = 0;

r = 12 ± √ (144 – 80);

r = 12 ± 8;

r1 = 4; r2 = 20.

Second solution:

(x – 20) ² + (y + 20) ² = 400.

Answer:

1) (x – 4) ² + (y + 4) ² = 16;

2) (x – 20) ² + (y + 20) ² = 400.