Equation of a tangent

deltaxray · #1 $Old$ October 26th, 2009, 02:45 AM

Hi everyone. I am having trouble with this question.

The tangent to a circle is prependicular to the radius at the point of contact.

Use this fact to show that the tangent to x^2 + y^2 = r^2 at the point (a,b)

has equation ax+by = r^2

Thanks

btw: can someone please tell me how to enter superscripts so i don't have to use the ^ sign. Thank you

Grandad · #2 $Old$ October 26th, 2009, 03:42 AM

Hello deltaxray

Quote:

Originally Posted by deltaxray $View Post$

Hi everyone. I am having trouble with this question.

The tangent to a circle is prependicular to the radius at the point of contact.

Use this fact to show that the tangent to x^2 + y^2 = r^2 at the point (a,b)

has equation ax+by = r^2

Thanks

btw: can someone please tell me how to enter superscripts so i don't have to use the ^ sign. Thank you

First how to avoid the use of ^ for superscripts.

Take the expression you wrote: x^2 + y^2 = r^2

Select it with the mouse (or <Shift>+<cursor keys>)

Press the $\Sigma$ button on the toolbar - this will wrap it round with [math] and [/math] tags.

Check the result using the 'Preview Post' button: $x^2 + y^2 = r^2$ .

The scripting language that does this is called LaTeX. There's lots of help on this forum for it. If you want to see any of the LaTeX code on what I've written below, just click it with the mouse.

Then:

The origin is the centre of the circle $x^2 + y^2 = r^2$ . So the gradient of the radius passing through $(a,b)$ is $\frac{\text{increase in }y}{\text{increase in }x}=\frac{b}{a}$ .

The tangent is perpendicular to this line; so its gradient is $-\frac{a}{b}$ .

So the equation of the tangent is:

$y-b=-\frac{a}{b}\Big(x-a\Big)$

i.e. $by-b^2=-ax+a^2$

or $ax+by = a^2+b^2$

$=r^2$ , since $(a,b)$ is a point on the circle $x^2+y^2=r^2$ .

Grandad

deltaxray · #3 $Old$ October 26th, 2009, 03:53 AM

Thanks grandad

$x^2$ i got it