Tangent to a circle where m=1

• May 20th 2009, 09:41 PM
ubhik
Tangent to a circle where m=1
Hi,

I have the equation of a circle, and hence I also know the centre and the radius. I need to find the tangent/s of gradient 1.

I have tried it many times by letting y=x + g the equation of the line and substituting it into the equation of the circle. Then solving using the quadratic formula, then solving for g.

I'm pretty sureI'm doing this wrong, because I get g=some surd, but I know it's going to be a neat answer, i.e. probably a whole number!

I've tried loads of times but I my method is wrong. Can anyone help?

Thank
• May 20th 2009, 10:05 PM
pickslides
for $y = \sqrt{1-x^2}$

find

$\frac{dy}{dx}= 1$
• May 21st 2009, 12:17 AM
Opalg
Quote:

Originally Posted by ubhik
Hi,

I have the equation of a circle, and hence I also know the centre and the radius. I need to find the tangent/s of gradient 1.

I have tried it many times by letting y=x + g the equation of the line and substituting it into the equation of the circle. Then solving using the quadratic formula, then solving for g.

I'm pretty sure I'm doing this wrong, because I get g=some surd, but I know it's going to be a neat answer, i.e. probably a whole number!

I've tried loads of times but I my method is wrong. Can anyone help?

Thank

Use the fact that the tangent is perpendicular to the radius. Suppose that the circle has centre (a,b) and radius r. If the tangent has gradient 1 then the radius will have gradient –1, so it will be in the direction of one of the vectors $\pm\tfrac1{\sqrt2}(r,-r)$. So the point where the tangent touches the circle will be $(a,b)\pm\tfrac1{\sqrt2}(r,-r) = (a\pm \tfrac r{\sqrt2},b\mp \tfrac r{\sqrt2})$. The equation of the line with gradient 1 through that point is then $y=x + (b-a\pm\sqrt2r)$. That gives you the equations of the two tangents having gradient 1.

Notice that if a, b and r are whole numbers then the equations of the tangents will also have whole numbers as their constant terms. Edit. That is wrong! (I forgot the factor $\sqrt2$ when I first wrote this. So if r is a whole number then the equations of the tangents will have a $\sqrt2$ in them.)
• May 21st 2009, 12:22 AM
ubhik
Quote:

Originally Posted by pickslides
for $y = \sqrt{1-x^2}$

find

$\frac{dy}{dx}= 1$

hi,

I think I understand what you mean.

But i don't think we can use calculus, and also I don't know how to differentiate : $x^2+y^2+12x-16y=0$ . We haven't been taught how to do it yet.

Quote:

Originally Posted by Opalg
Use the fact that the tangent is perpendicular to the radius. Suppose that the circle has centre (a,b) and radius r. If the tangent has gradient 1 then the radius will have gradient –1, so it will be in the direction of one of the vectors $\pm(r,-r)$. So the point where the tangent touches the circle will be $(a,b)\pm(r,-r) = (a\pm r,b\mp r)$. The equation of the line with gradient 1 through that point is then $y=x + (b-a\pm2r)$. That gives you the equations of the two tangents having gradient 1.

Notice that if a, b and r are whole numbers then the equations of the tangents will also have whole numbers as their constant terms.

Thanks Opalg,

I think this will help.