SOLUTION: Hi I need help answering these two questions. I have attempted both however my answer appears to be way off: i. Find the equation of a circle, whose center is at (6,1) and acts

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Question 1106436: Hi I need help answering these two questions. I have attempted both however my answer appears to be way off:
i. Find the equation of a circle, whose center is at (6,1) and acts as a tangent to the line: 3x-4y= 11
ii. The function f(x)= ax^2 + bx + c has roots (3,4) and it's graph passes through (5,6). Find a,b,c and decide if the graph f(x) opens up or down and find it's vertex.
Thank you so much, really appreciated <3
Found 2 solutions by greenestamps, josgarithmetic:
Answer by greenestamps(13200) (Show Source):
You can put this solution on YOUR website!

(1) You have the coordinates of the center; all you need is the radius.

The radius is the shortest distance from the center to the given line. The formula for the (shortest) distance from a given point (p,q) to a given line ax+by+c=0 is

You can finish from there.

(2) (note: "its", not "it's")

If the roots are 3 and 4, then the function is
where a is a constant to be determined.

You can find the constant by knowing that y is 6 when x is 5:

The function is

The function opens up. But you already knew that, since the graph passes through (3,0), (4,0), and (5,6).

Answer by josgarithmetic(39620) (Show Source):
You can put this solution on YOUR website!
Only here summarizing (i.):

Line perpendicular containing given center of circle,
slope -4/3,

Intersection of that with the given tangent line , is at ( 141/25, 37/25 ).

SQUARE of distance from (6,1) center and ( 141/25, 37/25 ) is

Equation of the circle wanted,