The equation of the tangent line to the circle passing through the point P is; y = (1/√3)x ± 2/√3
<h3>How to find the equation of the tangent?</h3>
I) We are given the equation of the circle as;
x² + y² = 1
Since angle of inclination is 30°, then slope is;
m = tan 30 = 1/√3
Then equation of the tangent will be;
y = (1/√3)x + c
Put (√3)x + c into the given circle equation to get;
x² + ((1/√3)x + c)² = 1
x² + ¹/₃x² + (2/√3)cx + c² = 1
⁴/₃x² + (2/√3)x + (c² - 1) = 0
Since we need to find value of c for equation to become tangent, then the above quadratic equation must have real and equal roots.
Thus;
((2/√3)c)² - 4(⁴/₃)(c² - 1) = 0
⁴/₃c² - ¹⁶/₃(c² - 1) = 0
⁴/₃c² - ¹⁶/₃c² + ¹⁶/₃ = 0
4c² = ¹⁶/₃
c² = ⁴/₃
c = √⁴/₃
c = ±²/√3
Thus, equation of tangent is;
y = (1/√3)x ± 2/√3
II) Radius from the given equation is 1. Thus, we will use trigonometric ratio to find the x and y intercept;
x-intercept is at y = 0;
0 = (1/√3)x ± 2/√3
-(1/√3)x = ±2/√3
Intercept is positive. Thus;
x = (2/√3)/(1/√3)
x = 1
y - intercept is positive at x = 0;
y = (1/√3)0 ± 2/√3
y = 2/√3
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, where r is the radius.