Math Is Fun Forum

I manage to solve it for the question below.

Sorry I don't know how to use the [math.][/math.] thing.

I tried with your method:
⇒ [y = m(x-10)+1] ⇒ x² + y² = 25
⇒ x² + (mx - 10m + 1)² - 25 = 0 ...... Expand and cleanup...
⇒ m²x² - 20m²x + 100m² +2mx - 20m + x² - 24 = 0 ...... Into quadratic equation.
⇒ (m² + 1)x² + (2m - 20m²)x + 100m² - 20m - 24 = 0 ...... Into The discriminant of a quadratic.
⇒ (2m - 20m²)² - 4(m² + 1)(100m² - 20m - 24) = 0 ...... Poof...!
⇒ 300m² - 80m - 96 = 0 ...... Solve with (-b±√(b² - 4ac))/(2a)...
⇒ m1 = (2+2√19)/15 and m2 = (2-2√19)/15 ...... Now we have the gradient, c = y - mx...
⇒ c1 = 1 - 10m1 and c2 = 1 - 10m2 ...... So...
⇒ c1 = (-1-4√19)/3 and c2 = (-1+4√19)/3 ....Subtitute into x² + y² = 25...
⇒ x² + (mx + c)² - 25 = 0 ...... Expand and cleanup...
⇒ (m² + 1)x² + 2mcx + c² - 25 = 0 ...... Solve with (-b±√(b² - 4ac))/(2a)...
⇒ (-(2mc) + √((2mc)² - 4(m² + 1)(c² - 25)))/(2(m² + 1)) ... THIS IS MADNESS...
⇒ x1 = 2.906821678 ...... Subtitude into y = mx + c and solve...
⇒ y1 = -4.068216776

So the first point of tangent is... approximately (2.9,-4.1)...
... There's no way I am gonna find the other point of tangent.
This is crazy. ( Took me 3 hours to solve the mistakes ).
At last I end up using online tools.

Okay, next, I tried using soroban's:
[math]⇒ (y - 1)/(x - 10) = -x/y ⇒
⇒ y² - y = -x² + 10x ⇒ 10x + y = x² + y² ⇒ y = 25 - 10x
⇒ x² + (25 - 10x)² = 25
⇒ x² + 625 - 500x + 100x² - 25 = 0 ⇒ 101x² - 500x + 600 = 0
Solving 101x² - 500x + 600 = 0 with (-b±√(b² - 4ac))/(2a)...
⇒ x1 = 2.906821678 x2 = 2.043673372 ... Am I seeing things?
Subtituting x1 and x2 into y = 25 - 10x...
⇒ y1 = -4.068216776 and y2 = 4.563266281

Resulting: (Sorry The image is over-sized...)

... x1 ≈ 2.906821678, y1 = -4.068216776 and x2 = 2.043673372, y2 = 4.563266281.
Wow... I solved this at once using soroban's method.
I'm sorry I misread the method, I thought it was something else.

Thank you Bobbym, Uh... Soroban, and Raghuram.
Thanks for a lot more than a dozen even more than a lot of a bunch!
Thank you! ( I don't know what else to say. )

Thanks, It helps a lot.
But in other words, that probably works only when the shape of the tangent is perpendicular.
But bobbym's method works on every points, and even tangents.
... except it needs a lot of hard work on expanding/cleaning up the equations.

Thanks a lot more than a bunch!