Math Is Fun Forum

FelizNYC wrote:

What is your comment here?

Sorry, it was just a reminder since, while you know it well, you used to miss applying it in your previous first solutions.

FelizNYC wrote:

the equation of the circle is x^2 + y^2 = r^2 and the equation of the tangent line is y = mx + b

Please note that the following simple steps don't need knowing the hint.

In general, when we have two traces (here, a circle and a straight line) and we like to find at which point(s) they may intersect, we simply solve their two equations.

We have:
x^2 + y^2 = r^2
y = mx + b
And their unknown variables are x and y.

By substituding y, we get:
x^2 + (mx + b)^2 = r^2
x^2 + m^2*x^2 + 2*m*x*b + b^2 = r^2
(1 + m^2)*x^2 + 2*m*b*x + (b^2 - r^2) = 0

This is a quadratic equation in the form of:
A*x^2 + B*x + C =0
where
A = (1 + m^2)
B = 2*m*b
C = (b^2 - r^2)

Its solution is:
x1 = [-B - sqrt(delta)]/(2*A)
x2 = [-B + sqrt(delta)]/(2*A)

Where delta is:
B^2 - 4*A*C

Therefore the delta of our equation above is:
= B^2 - 4*A*C
= (2*m*b)^2 - 4*(1 + m^2)*(b^2 - r^2)
= 4*m^2*b^2 - 4*(b^2 - r^2 + m^2*b^2 - m^2*r^2)
= 4*(m^2*b^2 - b^2 + r^2 - m^2*b^2 + m^2*r^2)
= 4*(- b^2 + r^2 + m^2*r^2)
= 4*[r^2*(1 + m^2) - b^2]

Since the straight line intersects the circle at one point only, x1 and x2 of this equation should be equal.
They are equal if its delta=0, that is:
4*[r^2*(1 + m^2) - b^2] = 0
r^2*(1 + m^2) - b^2 = 0
r^2*(1 + m^2) = b^2

FelizNYC wrote:

Bob wrote:

Both look good to me, well done.

Bob

Well, well. . .it's about time. I finally do something right, huh?

By the way, it is always about time for me too. Some problems that need innovative ways to solve took me many years to finish properly.

I mean if one needs 'really' to solve something, he will do it 'for sure' sooner or later.
Inversely, even the greatest genius won't solve a problem in which he is not interested.

FelizNYC wrote:

KerimF wrote:

You did well for r by applying the right equation of a circle. Then... I didn't see it.

I guess you know that:
x^2 + y^2 = r^2
means that the circle center is at (0, 0).

Is my answer right?

But... your exercise is to find the standard form of the equation of the circle with center (-3, 1) ... not (0,0)

If you substitute ( x , mx+c) into the circle equation you get that hint.

A quadratic Ax^2 + Bx + C = 0 has only one solution if B^2 - 4AC is zero.  Part 1 comes from that equation substituting for A, B and C. (Don't mix up C and c here; they're not the same.)

If P is the point of tangency and origin O then OP is perpendicular to the tangent so it has equation y = -x/m

Sub into the equation for the circle to find x and then y.

What is the question for part c?

Bob

x may be anything but y must always be 2.

Bob

Correct.

B

You haven't allowed for any clearance at the bottom with the ground.

The maximum height is 165; subtract the diameter and the lowest point must be 15m off the ground.

So the centre is at 15 + 75 up the y axis.

Bob

You keep doing the hard bit ok then going back to the wrong circle equation.

You need (x-g)^2 + (y-h)^2 = r^2 for all the questions.

Bob

You did well for r by applying the right equation of a circle. Then... I didn't see it.

I guess you know that:
x^2 + y^2 = r^2
means that the circle center is at (0, 0).