Çözüldü Implicit Differentiation - Coordinate Geometry

What is the y-intercept of the tangent line to the point P(1, 2) on the circle x^2 + y^2 - 2x + 4y - 11 = 0 ?

Solution with Differentiation:
2x + 2yy' - 2 + 4y' = 0 ⇒ y' = f(x, y) = (1 - x) / (2 + y) and the slope of the tangent m = f(1, 2) = (1 - 1) / (2 + 2) = 0
So the equation of the tangent line is y - 2 = 0(x - 1) ⇒ y = 2
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Solution without Differentiation:
Let the equation of the tangent line be y = mx + n
Intersection of the circle with the line y gives the equation;
x^2 + (mx + n)^2 - 2x + 4mx + 4n - 11 = 0 and after organizing it into a quadratic equation;
(1 + m^2)(x^2) + 2(mn + 2m - 1)x + n^2 + 4n - 11 = 0
Because of the tangential situation, by equating the reduced discriminant to zero;
(mn + 2m - 1)^2 - (1 + m^2)(n^2 + 4n - 11) = 0....(I)
The line passing the point P(1, 2) gives the equation 2 = m + n ⇒ n = 2 - m....(II)
Substituting n in (II) to (I) yields [ m(2 - m) + 2m - 1 ]^2 - (1 + m^2)[ (2 - m)^2 + 4(2 - m) - 11 ] = 0 and after expanding, leaving the simplification works to the students who may be interested;
16m^2 = 0 ⇒ m = 0 ⇒ n = 2 - 0 = 2
So the equation of the tangent line is y = 0x + 2 = 2.
That is, the y-intercept is 2.

Original Question:
https://i.ibb.co/VHF1v1G/circle-tangent.png
https://scontent-mxp1-1.xx.fbcdn.ne...=a8ad5cc450104dad1aed89a3a503c252&oe=5D770F5A
https://www.facebook.com/photo.php?fbid=196555151321913&set=g.703383433087490&type=1&theater&ifg=1
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Note for the teachers who might wish to use this question on their own exams:
Let the student check and find on her / his own whether the point P is on the circle or not; so, by hiding that differentiation could be applied, rewording of the question can be:
What is the y-intercept of the line tangent to the circle x^2 + y^2 - 2x + 4y - 11 = 0 and passing through the point P(2, 1)?