Question

Circle Geometry Question?

A tangent is drawn from point P(-2,2) to the circle `x^2+y^2-6x-2y=0` at T. Find the length of PT and find the gradient of the tangent from P to the circle.

Answer 1

PT = 4

Explanation:

!

Answer 2

`color(blue)("gradient"_1=(-5+4sqrt(10))/(15)`

`color(blue)("gradient"_2=(-5-4sqrt(10))/(15)`

`color(blue)("Distance " PT = 4)`

Explanation:

First rearrange the circle equation to get the form:

`(x-h)^2+(y-k)^2=r^2`

`:.`

`(x-3)^2+(y-1)^2=10`

Now, we require the tangent line to pass through `(-2,2)`

Using point slope form of a line with gradient k:

`y-2=k(x+2) \ \ \[1]`

`y=k(x+2)+2`

This line has to touch the circle so we substitute this line in the circle equation. As will be seen there will in fact be two tangent lines. The equation after substitution will be quadratic and since we only want to touch the circle, we will need to find values of k such that the quadratic has repeated roots. ie, the discriminant will be zero.

Substituting:

`(x-3)^2+( k(x+2)+2-1)^2-10=0`

Expand:

`x^2-6x+k^2x^2+4k^2x+2kx+4k^2+4k`

Rearrange this into the form: `ax^2+bx+c`

`(1+k^2)x^2+(-6+4k^2+2k)x+(4k^2+4k)`

Using the discriminant, and equating to zero:

`b^2-4ac=0`

`(-6+4k^2+2k)^2-(4(1+k^2)(4k^2+4k))=0`

Expanding `LHS` and simplifying:

`36-60k^2-40k=0`

`60k^2+40k-36=0`

Solving using the quadratic formula:

`k=(-40+-sqrt((40)^2-4*(60)*(-36)))/(120)`

`k=(-40+-sqrt(10240))/(120)=(-40+-32sqrt(10))/(120)`

`=(-5+-4sqrt(10))/(15)`

So our two gradients are:

`k=(-5+4sqrt(10))/(15) and k=(-5-4sqrt(10))/(15)`

Using point `(-2,2)` and point slope form of a line:

`y-2=(-5+4sqrt(10))/(15)(x+2)`

`y=(-5+4sqrt(10))/(15)x+2((-5+4sqrt(10))/(15))+2`

`color(blue)(y=(-5+4sqrt(10))/(15)x+((20+8sqrt(10))/15)`

And for the other value of k we have:

`color(blue)(y=(-5-4sqrt(10))/(15)x+((20-8sqrt(10))/15)`

Now we now the equation of the line, we need to find the coordinates at the point of tangency.

Using our equation for the tangent line, we substitute this in the circle equation.

`(x-3)^2+((-5+4sqrt(10))/(15)x+((20+8sqrt(10))/15)-1)^2-10=0`

This gives:

`x=(2(19+sqrt(10))(41+4sqrt(10)))/1521`

`y=(2(1+37sqrt(10))(1841+328sqrt(10)))/177957`

This is only for one point, but since the distances are the same for both points and the big calculations for finding these, I won't do the other point.

The distance form P to T can be found using the distance formula:

`|d|=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`|d|=sqrt((-2-x)^2+(2-y)^2)`

Believe it or not this is just `4`

So gradients are:

`m_1=(-5+4sqrt(10))/(15)`

`m_2=(-5-4sqrt(10))/(15)`

Distance PT is:

`4`

PLOT:

A lot of the calculations were done using a computer algebra system due to the size of them. I feel sure there is an easier method than this, and hopefully someone else will give an answer that is a bit more elegant than this one, but maybe it will help you.