# The line 4x-3y-40=0 touches the circle (x-2)^2+(y-6)^2=100 at point P. Show that the radius at P ia prependicular to the line.

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To solve this , we need to find the points were the 2 lines meet in order to find the gradient of the perpendicular straight line which is perpendicular to 4x-3y-40=0.

perpendicular lines have certain properties and we know that where `m` is the gradient or slope `m_(1) times m_(2) = -1`

So we find the gradient of the existing straight line `4x-3y-40 =0`

by rearranging the equation in standard form:`y=mx+c` `therefore 4x-40 = 3y`

`therefore 3y = 4x-40`

`therefore y=4/3 x -40/3`

So we need to prove that the gradient of the perpendicular line

`4/3 times m_(2) = -1`

`therefore m_(2) = -1 times 3/4 = -3/4`

We have one point on the perp line (2;6) which we saw from the circle equation :`(x-2)^2 + (y-6)^2 = 100` where the centre is at x=2 and y=6 .

We also know that the radius = 10 because our circle equation represents `r^2` because `x^2 +y^2 = r^2` `therefore r=sqrt100` and can only be positive becasue it is a distance.

Use distance formula to find the other point (where the 2 lines meet). Because there are 2 unknowns you will have to substitute and it is easiest to substitute `y= 4/3 x - 40/3` into the formula. Remember the two lines meet at this point which is what enables us to substitute one in to the other as they are equal at **that** point.

`d = sqrt((x_(2) - x_(1))^2 + (y_(2) - y_(1))^2)`

we know that the distance is 10 (the radius) and we have one of the x values and one of the y values (2;6). For the other y value we will substitute our other equation as previously mentioned:

`10= sqrt((2-x)^2 + (6-(4/3x- 40/3))^2)`

Now we only have one unknown value to find

`10 = sqrt((4-4x+x^2 + (6+40/3 - 4/3x)^2)`

the 6 is added to the 40/3 to get 19 1/3 or 58 /3 which is then squared as is the rest of this bracket `(58/3 - 4/3x)^2`

`10= sqrt(x^2 -4x +4 + 3364/9 - (464x)/9 + 16/9x^2`

now put the like terms together (ie `x^2 + 16/9 x^2` etc)

`10=sqrt(25/9 x^2 - 500/9 x+ 3400/9)`

Square both sides to remove the square root and multiply everything by 9 to remove the denominator:

`100 times 9= 25x^2 - 500x + 3400`

Now divide everything by 25 (common factor on the right hand side)

`900/ 25 = x^2 - 20x +136`

Now bring everything to one side (900 divide by 25 = 36):

`x^2 - 20x +136 - 36 = 0`

`therefore x^2-20x +100 = 0`

simplify

`therefore (x-10)^2=0`

`therefore x=10`

we need to find y at the same point so substitute x into one of your original equations:

`4(10) - 3y -40 = 0`

`-3y=0`

`therefore y=0`

So now we have 2 points for our perpendicular straight line:

`y=mx+c` becomes at point (2;6)`6=m(2) + c` and at point (10;0)`0=m(10) +c`

Now we use simultaneous equations to solve. As they are both linear (straight lines) we can subtract one from the other:

`6=2m+c`

`-(0=10m+c)`

Note how c cancels out and we are left with 6-0 and 2m-10m

`therefore 6=-8m`

`therefore 6/-8 = m`

Simplify :

`therefore -3/4 = m`

Note how we needed to show that our lines were perpendicular and :

`4/3 times -3/4 =-1` has been proven

**Therefore the radius is perpendicular to the line `4x-3y-40=0`**

** at point P**

The line 4x-3y-40=0 touches the circle (x-2)^2+(y-6)^2=100 at point P. Show that the radius at P ia prependicular to the line.

The line touches cirle at P. So let find cordinate of pont P.

4x-3y=40

4x=40+3y

x=(40+3y)/4 , (i)

substitute in equation of cirle.

((40+3y)/4-2)^2+(y-6)^2=100

1024+9y^2+192y+16y^2+576-192y=1600

16y^2=0

y=0

substitute y=0 in (i) ,

x=10

Thus coordinate of P=(10,0)

Centre of circle O =(2,6)

slope of OP=(0-6)/(10-2)=-3/4

slope of the given line 4x-3y=40 is

-3y=-4x+40

y=(4/3)x-40/3

thus slope of given line is (4/3)

product of slop of OP and given line= (-3/4)(4/3)=-1

**Thus OP ,radius is perpendicular to given line at P .**