1. ## Proving a result.

I've come across this question and was wondering how to prove it, particularly since the book from which the question has come from assumes no knowledge of calculus.

The equation of a straight line is y=ax+b, where a and b are constants.

The equation of a circle is: x^2 + y^ = 64.

The straight line is a tangent to the circle.

Prove that a^2 + 1 = b^2 / 64

Any ideas?

2. Originally Posted by Cairo
I've come across this question and was wondering how to prove it, particularly since the book from which the question has come from assumes no knowledge of calculus.

The equation of a straight line is y=ax+b, where a and b are constants.

The equation of a circle is: x^2 + y^ = 64.

The straight line is a tangent to the circle.

Prove that a^2 + 1 = b^2 / 64

Any ideas?
There is only one solution to the simultaneous equations:

$\displaystyle x^2 + y^2 = 64$ .... (1)

$\displaystyle y = ax + b$ .... (2)

Therefore the quadratic equation $\displaystyle x^2 + (ax + b)^2 = 64$ has only one solution. So equate the discriminant to zero and see what happens.

3. Originally Posted by Cairo
I've come across this question and was wondering how to prove it, particularly since the book from which the question has come from assumes no knowledge of calculus.

The equation of a straight line is y=ax+b, where a and b are constants.

The equation of a circle is: x^2 + y^ = 64.

The straight line is a tangent to the circle.

Prove that a^2 + 1 = b^2 / 64

Any ideas?

Ast there is one unique point of intersection between both curves, we get a unique solution to the quadratic:

$\displaystyle x^2+(ax+b)^2=64\Longrightarrow (a^2+1)x^2+2abx +b^2-64=0\Longrightarrow \Delta=4a^2b^2-4(a^2+1)(b^2-64)=0$ (why?)

$\displaystyle \Longrightarrow 256a^2-4b^2+256=0$...and deduce the result.

Tonio

4. Originally Posted by Cairo
I've come across this question and was wondering how to prove it, particularly since the book from which the question has come from assumes no knowledge of calculus.

The equation of a straight line is y=ax+b, where a and b are constants.

The equation of a circle is: x^2 + y^ = 64.

The straight line is a tangent to the circle.

Prove that a^2 + 1 = b^2 / 64

Any ideas?
1. Calculate the coordinates of the points of intersection between the straight line and the circle. Plug in the term of y into the equation of the circle and solve for x.

2. The intersecting line becomes a tangent if the radical equals zero. And exactly there you'll find the given condition.

5. Thanks for this.

I've got the result now.