Question

If the length of the tangent from (5, 4) to the circle
x² + y² + 2ky = 0
is 1, then find k.​

Answer 1

Answer:

x^2+y^2+2ky=0

at the point (5,4)

5^2+4^2+2k(4)=0

25+16+8k=0

41+8k=0

8k=-41

k=-41/8

Answer 2

Answer:

$k=-5$

Tangent of a circle:

Tangents to circles are lines that cross the circle at a single point. Point of tangency refers to the location where a tangent and a circle converge. The circle's radius, where the tangent crosses it, is perpendicular to the tangent. Any curved form can be considered a tangent. Tangent has an equation since it is a line. In order to better comprehend the idea, we will solve an example while simultaneously discussing the general equation of a tangent in slope form.

Step-by-step explanation:

When connected to the circle's centre (0,k), the radius, the tangent from (5,4), and the point (5,4) create a right-angled triangle.

$\therefore (length \;of\; tangent)^2 + (radius)^2 = (distance \;of \;(5,4) \;from \;center \;of \;the \;circle)^2$

$\therefore (length \;of\; tangent) ^2 +k ^2 =(5-0) ^2 +(4+k) ^2$

$\therefore 1+k ^2 =25+k ^2 +8k+16\\\therefore 8k+40=0\\k=-5$

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