If the line x + y = C is tangent to the circle X & # 178; + Y & # 178; = 2, then C = () 3Q

If the line x + y = C is tangent to the circle X & # 178; + Y & # 178; = 2, then C = () 3Q

The distance d between the center of the circle (0, 0) and the straight line x + Y - C = 0 is equal to the radius √ 2
d = |-c|/√2 = √2
|c|= √2
c = ±√2

Find the tangent equation passing through point a (1, - 7) and tangent to the circle x square + y square = 25,

Let the tangent equation be y = K (x-1) - 7, that is, kx-y-k-7 = 0, the center of the circle is known to be (0,0), and the radius r = 5. Because the distance from the center of the circle to the tangent is equal to the radius, so r = i-k-7i / √ (k ^ 2 + 1) = 5 square, K ^ 2 + 14K + 49 = 25K ^ 2 + 2512k ^ 2-7k-12 = 0 (4K + 3) (3k-4) = 0k = - 3 / 4 or 4 / 3, the obtained equation is 4x-3y-25 = 0 or 3x + 4Y + 25 = 0

Find the tangent equation passing through point (4,0) and tangent to the square of circle (x-1) + (y-4) = 25

Let the linear equation over (4,0) be y = K (x-4)
When the equation about X has only one root, K is the slope of the straight line
y=k(x-4)
y-4=kx-4k-4
(x-1)^2 + (kx-4k-4)^2 = 25
When Δ = 0, there are two solutions,