Locus - Tangent to 2 circles (1 Viewer)

[davidbarnes]

Ok, so heres another locus question I'm having issues with.

"Show that the circles X^2 + Y^2 = 4and X^2 + 2x + y^2 - 4y - 4 = 0 both have
3x + 4y + 10 = 0 as a tangent."

So with X^2 + 2x + y^2 - 4y - 4 = 0, centre = (-1, 2), r = 2
and with X^2 + Y^2 = 4, centre = (0, 0), r = 2 right?

Thats where I'm confused and not sure how to go about next.

 

[ssglain]

I'm not so sure that this is a locus question.

As a hint - remember that one way to define a tangent to a circle is that the tangent is a line which has only ONE point of contact with the circle. This will mean that when the equations of the line and the circle are solved simulaneously, there is only ONE real solution. Now bells should be ringing about the discriminant. Try for yourself. The solution is under the spoiler.

Spoiler

First, x² + y² = 4 & 3x + 4y + 10 = 0

From the line:
y = (1/4)*(10 - 3x)

Put this in the circle:
x² + [(1/4)*(10 - 3x)]² = 4
x² + (1/16)*(100 - 60x + 9x²) = 4
16x² + 100 - 60x + 9x² = 64
25x² - 60x + 36 = 0

discriminant = b² - 4ac = (-60)² - 4(25)(36) = 3600 - 3600 = 0 --> only ONE real solution
.: 3x + 4y + 10 = 0 is a tangent to x² + y² = 4 since there is only ONE point of contact.

Second, x² + 2x + y² - 4y - 4 = 0 & 3x + 4y + 10 = 0
Similarly show that discriminant = 0

You can also do this question by considering another tangent definition: the tangent to a circle is perpendicular to the radius at the point of contact. You can show that the line is a tangent by showing that the perpendicular distance between the line and the each circle equals to the length of the radius. I don't have time to write out working for this method, but give this a go yourself.

 

[beentherdunthat]

I'm not so sure that this is a locus question.

As a hint - remember that one way to define a tangent to a circle is that the tangent is a line which has only ONE point of contact with the circle. This will mean that when the equations of the line and the circle are solved simulaneously, there is only ONE real solution.

Yep, thats about it... They ask this question a lot, I used to get confused about it initially as well. But it's really simple

You can also do this question by considering another tangent definition: the tangent to a circle is perpendicular to the radius at the point of contact.

If your doing 2unit maths only, then stick with the top method. This is extra info. from 3unit

 

[davidbarnes]

I'm not so sure that this is a locus question.

As a hint - remember that one way to define a tangent to a circle is that the tangent is a line which has only ONE point of contact with the circle. This will mean that when the equations of the line and the circle are solved simulaneously, there is only ONE real solution. Now bells should be ringing about the discriminant. Try for yourself. The solution is under the spoiler.

You can also do this question by considering another tangent definition: the tangent to a circle is perpendicular to the radius at the point of contact. You can show that the line is a tangent by showing that the perpendicular distance between the line and the each circle equals to the length of the radius. I don't have time to write out working for this method, but give this a go yourself.

Thanks for tee help. I follwoed most of that, althoghu don't quite get what formula/means you used to get 3x + 4y + 10 = 0 into the form y = (1/4)*(10 - 3x), such as why is 3x last in the equation, etc?

 

[SoulSearcher]

Thanks for tee help. I follwoed most of that, althoghu don't quite get what formula/means you used to get 3x + 4y + 10 = 0 into the form y = (1/4)*(10 - 3x), such as why is 3x last in the equation, etc?

3x + 4y + 10 = 0
4y = -(10 + 3x)
y = (-1/4)*(10 + 3x)

 

[ssglain]

3x + 4y + 10 = 0
4y = -(10 + 3x)
y = (-1/4)*(10 + 3x)

Oops my bad again. LOL. I thought the line was 3x + 4y = 10. These silly mistakes are going to cost me a lot in the HSC. Dunno what I'd do without you.