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February 3rd, 2018, 10:53 PM   #1
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Tangents to Circle

Hi all

Can someone help me to find the co-ordinates of point P please?

I understand that the tangents are the same length (because they're both coming from a single external point).

Does this mean an isosceles triangle is involved in solving it?

Any pointers on the next step would be appreciated.

Thanks

Jim
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February 3rd, 2018, 11:06 PM   #2
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It should be pretty clear that

i) $B=(0,-5)$

ii) $P=(P_x,-5)$

so we just need to solve for $P_x$

The tangent line at $(3,4)$ will have slope $-\dfrac{3}{4}$ (show this)

The slope of this line is also $\dfrac{4-(-5)}{3-P_x}= \dfrac{9}{3-P_x}$

$\dfrac{9}{3-P_x} = -\dfrac{3}{4}$

$P_x = 15$
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February 3rd, 2018, 11:41 PM   #3
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Hi Romsek

That is a brilliant answer which I understand.

The only part that I don't understand is how you work out the gradient of the tangent?

What is the simplest way of doing that please?

Regards

Jim
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February 4th, 2018, 12:08 AM   #4
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Originally Posted by Jimbo77 View Post
Hi Romsek

That is a brilliant answer which I understand.

The only part that I don't understand is how you work out the gradient of the tangent?

What is the simplest way of doing that please?

Regards

Jim
What do you know about the tangent line of a circle with respect to the radius of the circle? Use this.
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February 4th, 2018, 02:14 AM   #5
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Or use the distance formula and the fact that the tangents are of equal length.

$$(x-3)^2+81=x^2\implies x=15$$
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February 4th, 2018, 06:10 AM   #6
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What do you know about the tangent line of a circle with respect to the radius of the circle? Use this.
I know it is always at right angles? Is this correct?
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February 4th, 2018, 06:13 AM   #7
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Or use the distance formula and the fact that the tangents are of equal length.

$$(x-3)^2+81=x^2\implies x=15$$
Sorry could you explain this a bit more please?
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February 4th, 2018, 11:48 AM   #8
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Distance formula.
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February 4th, 2018, 11:59 AM   #9
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