Normally, you have a normal to the circle (x-a)² + (y-b)² = r² at a point (x₀,y₀) on the circle.
The normal line passing through the points (a,b) (because the normal of a circle always passes through the centre) and (x₀,y₀) is (b-y₀)x - (a-x₀)y = bx₀ - ay₀.
This is perpendicular to the tangent line to the circle (x-a)² + (y-b)² = r² at the tangent point (x₀,y₀) which is (x₀-a)(x-a) + (y₀-b)(y-b) = r², or (x₀-a)x + (y₀-b)y = r² + a(x₀-a) + b(y₀-b) in the same form as the normal above.
A line that's normal to a circle must pass through the center, and any such line works. The center of this circle is (1,2), and an arbitrary line through that point can be constructed with
a(x - 1) + b(y - 2) = 0 . . . . where a and b are not both zero
That's obviously an linear equation in x and y. Further, it's obviously true when x=1 and y=2 so the line includes the point (1,2).
You could also use the point-slope formula, but that wouldn't include the vertical line through (1,2). The above simplifies in to standard form:
Answers & Comments
(x-1)² + (y-2)² = 10
y = mx + a√1+m²
x²-2x+1+y²-4y+4 = 10
x²+y²-2x-4y-5 = 0
tangent is S₁ = 0
xx₁ + yy₁ - (x+x₁) – 2(y+y₁)-5 = 0
xx₁ + yy₁ - x-x₁ - 2y – 2y₁ - 5 = 0
tangent perpendicular is normal
x₁(x-1) + y₁(y-2) + (-x-2y-5) = 0
Normal (y-2)x₁-(x-1)y₁+k = 0.
Normally, you have a normal to the circle (x-a)² + (y-b)² = r² at a point (x₀,y₀) on the circle.
The normal line passing through the points (a,b) (because the normal of a circle always passes through the centre) and (x₀,y₀) is (b-y₀)x - (a-x₀)y = bx₀ - ay₀.
This is perpendicular to the tangent line to the circle (x-a)² + (y-b)² = r² at the tangent point (x₀,y₀) which is (x₀-a)(x-a) + (y₀-b)(y-b) = r², or (x₀-a)x + (y₀-b)y = r² + a(x₀-a) + b(y₀-b) in the same form as the normal above.
Animated Graph: https://www.desmos.com/calculator/tedrnmmfrt
A line that's normal to a circle must pass through the center, and any such line works. The center of this circle is (1,2), and an arbitrary line through that point can be constructed with
a(x - 1) + b(y - 2) = 0 . . . . where a and b are not both zero
That's obviously an linear equation in x and y. Further, it's obviously true when x=1 and y=2 so the line includes the point (1,2).
You could also use the point-slope formula, but that wouldn't include the vertical line through (1,2). The above simplifies in to standard form:
ax + by = a + 2b
A circle has an infinite number of lines that are normal to the circle. Any line through the center is a normal.
Your circle has a center of (1,2) and a radius of √10. Any line through the center is a normal so take your pick.
Example:
y = x + 1
y = -x + 3