Question

5. What is the equation of the tangent line to the circle x^2+y^2=1 through the point (6,0)?

Answer 1

Answer:

There is no tangent line of the given circle at (6, 0).

Step-by-step explanation:

Given equation of the circle,

$x^2 + y^2 = 1$

∵ equation of a circle is $(x-h)^2 +(y-k)^2 = r^2$,

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, $6^2 + y^2 = 1$

$36 + y^2 = 1$

$y^2 = 1- 36$

$y= i\sqrt{35}\neq 0$

i.e., (6, 0) does not lie on the circle,

Hence, there is no tangent line of the given circle at (6, 0).