Answer:
There is no tangent line of the given circle at (6, 0).
Step-by-step explanation:
Given equation of the circle,
[tex]x^2 + y^2 = 1[/tex]
∵ equation of a circle is [tex](x-h)^2 +(y-k)^2 = r^2[/tex],
Where, (h, k) is the center of the circle and r is the radius,
By comparing,
Center of the given circle = (0, 0),
Radius of the circle = 1 unit
Now, check whether point (6, 0) lie on the circle,
if x = 6, [tex]6^2 + y^2 = 1[/tex]
[tex]36 + y^2 = 1[/tex]
[tex]y^2 = 1- 36[/tex]
[tex]y= i\sqrt{35}\neq 0[/tex]
i.e., (6, 0) does not lie on the circle,
Hence, there is no tangent line of the given circle at (6, 0).
