If the distance between the vertex and the origin of the image of quadratic function y = AX2 + BX + C is 5, then C =?

If the distance between the vertex and the origin of the image of quadratic function y = AX2 + BX + C is 5, then C =?

It is known that the coordinates of the three vertices of the triangle ABC are a (- 2,3), B (1,2), C (5,4) BA = (- 3,1) BC = (4,2)
The process should be detailed

Vector Ba = (- 3,1). Vector BC = (4,2)
BA.BC= (-3)*4+1*2=-10.
|BA|=√[(-3)^2+1^2]=√10.
|BC|=√(4^2+2^2)=2√5
cosB= BA.BC/ |BA||BC|
=(-10)/2√5*√10.
=-√2/2.
∴∠B=135°

1. In △ ABC, ∠ B = ∠ C = α, (0 ° < α)

（1）∠1=150°－β,∠2=30°+β－α；………………………………… (1 '× 2 = 2') (2) from β = ∠ 2 or ∠ 1 = ∠ CQP, α = 30 degree is obtained ..………… ..… .… When β changes within the allowable range, α = 30 ° always has △ ABP ∽ PCQ. ② from β = ∠ 1 or ∠ 2 = ∠ CQP, β = 75 ° ..…………… ...… When α changes within the allowable range, β = 75 ° there is always △ ABP ∽ QCP

Point a is a fixed point, line BC slides on the fixed line L. given that the modulus of BC is 4, the distance between point a and line L is 3, the trajectory equation of the outer center of triangle ABC is obtained

Let a = (0,3), B = (T, 0), C = (T + 4,0), the equation of the vertical bisector L1 of BC is x = t + 2, the equation of the vertical bisector L2 of AB is tx-3y = (T ^ 2-9) / 2, and the intersection of L1 and L2 q = (T + 2, T ^ 2 / 6 + 2T / 3 + 3 / 2) is the outer center of triangle ABC