Verified answer

The equation of tangent at the point (h, k) on the circle

x² + y² +2gx +2fy + c= 0 is

hx + ky + g(x + h) + f(y + k) + c = 0

=> (h+g)x + (k+f)y +gh +fk + c = 0

If this tangent line is also a tangent to the circle

x² + y² = r², then the perpendicular distance from the center of the circle (0, 0) to this line should be equal to the radius r of the circle.

=> l gh + fk + c l / √[(h+g)² + (k+f)²] = r

=> (gh + fk + c)² = r² (h² + k² + 2gh + 2fk + g² + f²) ... (1)

Now, as (h, k) is on the circle x² + y² +2gx +2fy + c= 0,

h² + k² + 2gh + 2fk + c = 0

=> h² + k² + 2gh + 2fk = - c ... (2)

Using the relation of (2) in (1)

=> (gh + fk + c)² = r² (g² + f² - c).

x^2 + y^2 + 2gx + 2fy + c = 0

x^2 + y^2 = r^2

Prove that:

(gh + fk + c)^2= r^2(g^2 + f^2 - c)

Put x=h and y=k into the first equation and use the second:

r^2 + 2gh + 2fk + c = 0

r^2 + (gh + fk + c) + (gh + fk) = 0

r^2 + (gh + fk) = (gh + fk + c)

[r^2 + (gh + fk) ]^2= (gh + fk + c)^2

The right side is good so lets just look at the left side:

Need to show:

[r^2 + (gh + fk) ]^2 = r^2(g^2 + f^2 - c)

[r + (gh + fk)/r]^ 2 = g^2 + f^2 - c

r^2 + 2(gh + fk) + (gh + fk)^2/r^2 = g^2 + f^2 - c

but: r^2 + 2(gh + fk) = -c ....... from the first circle

leaving:

(gh + fk)^2/r^2 = g^2 + f^2

(gh + fk)^2 = (g^2 + f^2) r^2 = (g^2 + f^2)(h^2 + k^2)

g^2h^2 + f^2k^2 + 2ghfk = g^2h^2 + f^2k^2 + h^2f^2 + g^2k^2

2ghfk = h^2f^2 + g^2k^2

h^2f^2 + g^2k^2 - 2ghfk = 0

(hf - gk)^2 = 0

So we need to show that hf - gk = 0 and this can be done by looking at the slope of the tangents. The tangent is the same for both circles so the slope must be the same.

Slope at a point (x,y) on a circle is -(x/y) or, in other words, -(x distance from center)/(y distance from center) for the point in question.

The first circle can be rewritten as:

(x + g)^2 + (y + f)^2 = something

So the center of the first circle is (-g,-f) and the slope of the tangent to the first circle is:

m1 = -(g + h)/(f + k)

This must equal the slope of the tangent from the other circle:

m2 = (-h/k)

(g + h)k = (f + k)h

gk + hk = fh + kh

hf - gk = 0 Which completes the proof

Note:

Slope of tangent for first circle:

2x+ 2y(dy/dx) + 2g + 2f(dy/dx) = 0

(dy/dx)(y + f) = -(x + g)

at x = h and y = k

(dy/dx) = -(h + g)/(k + f)

"c" isn't the radius of the circle. in spite of the indisputable fact that the value of the radius might nicely be chanced on from a records of the values of "c", "g" and "f" by utilizing utilizing a technique noted as winding up the sq..