Tangent equation of P (- 5, - 2) and circle x ^ 2 + y ^ 2 = 25 RT

Tangent equation of P (- 5, - 2) and circle x ^ 2 + y ^ 2 = 25 RT

Let the tangent l equation be: y = KX + B ∵ line L passing through point P (- 5, - 2) ∵ - 2 = - 5K + B, that is, B = 5k-2 ∵ kx-y + 5k-2 = 0, line L is tangent to circle X & sup2; + Y & sup2; = 25 ∵ d = | 5k-2 | / (radical K & sup2; + 1) = 5 ∵ k = - 21 / 20, then the tangent l equation is: 21x + 20Y + 145 = 0

Tangent equation passing through point (1, - 7) and tangent to circle x2 + y2 = 25______ ．

If the slope of the tangent does not exist, because the tangent passes through the point (1, - 7), the equation of the straight line is x = 1 and the circle x2 + y2 = 25, which does not meet the requirements. If the slope of the tangent exists, let the slope of the tangent be K, because the tangent passes through the point (1, - 7), let the equation of the tangent be y + 7 = K (x + 1), that is, kx-y + k-7 = 0

Given the circle x ^ 2 + y ^ 2 = 25, the tangent equation of B (- 5,2) is solved

Obviously, x = - 5 is a tangent of the circle
When the tangent slope exists, let the tangent equation be Y-2 = K (x + 5), that is, y = KX + 5K + 2
If the distance from the center of the circle to the straight line = radius, there is:
|5K + 2 | / radical (k ^ 2 + 1) = 5
25k^2+20k+4=25k^2+25
k=21/20
The other line is y = 21 / 20x + 7.25