Quad. Function - Question.. :) (1 Viewer)

[Smile12345]

Hello All.

Could you please help me with a couple of questions... I seem to be stumped...

Q. Is this line (see below) a tangent to the circle x^2 + y^2 = 4
Line: 4x + 3y -10 = 0

Q. Find values of a, b, and c for which: x^2 + 4x -3 is congruent to (symbol with 3 - on top of each other) a(x+1)^2 + b(x+1) + c

Thanks heaps in advance.

 

[asianese]

What is the quadratic equation whose roots are the intersections of x^2 + y^2 = 4 and 4x + 3y - 10 = 0? What properties of this quadratic tell you if the line is a tangent to the circle?

Expand the right hand side.

 

[SharkeyBoy]

For your first question, you need to do simultaneous equations. So first, make x or y the subject of the equation.
So x = (10 - 3y) / 4
or y = (10 - 4x) / 3
therefore, if we substitute x: [(10-3y)/4]^2 + y^2 = 4
Then solve to see what y will equal and then substitute in to see if it equals up

For your second question, I like to do these type of questions by expanding the right hand side first, meaning it would equal:
ax^2 + 2ax + a + bx + b + c
= x^2(a) + x(2a + b) + (a + b +c)
then we need to equate this to the LHS
coefficient of x^2 in x^2 + 4x -3 is 1
therefore, a = 1
coefficient of x in x^2 + 4x -3 is 4
therefore 2a + b = 4
since a =1, then b = 2
constant in x^2 + 4x -3 is -3
therefore, a + b + c = -3
since a = 1, b = 2, then c = -6
that should be right

 

[Smile12345]

Thanks .... Q2 - Yeah that's right thanks.

 

[Drongoski]

Or:

for Q1 you can find perpendicular distance of centre of the circle, viz (0,0) from the line 4x+3y-10 = 0. you will find this to be equal to 2, exactly the radius of the circle. Therefore the given line is tangent to the circle.

 

[Smile12345]

Or:

for Q1 you can find perpendicular distance of centre of the circle, viz (0,0) from the line 4x+3y-10 = 0. you will find this to be equal to 2, exactly the radius of the circle. Therefore the given line is tangent to the circle.

Thanks heaps.... That is an awesome way to do it.