# Find the equation of the tangent to each circle

• May 24th 2009, 10:01 AM
Maths_noob
Find the equation of the tangent to each circle
Find the equation of the tangent to each circle at the point specified

1 Circle $x^2 + y^2 - 8x - 42y =$ 0, point (8, 5)

2 Circle $x^2 + y^2 - 5x - 9y - 16 = 0$ point(6, -4)

Any help would be great, don't even know what I should read up on to actually find out how to do it either... (Crying)

Thanks
• May 24th 2009, 10:28 AM
Soroban
Hello, Maths_noob!

Please check the problem for typos.

The given points are NOT on their respective circles.

Quote:

Find the equation of the tangent to each circle at the point specified:

$(1)\;x^2 + y^2 - 8x - 42y \:=\:0,\quad(8, 5)$

$(2)\;x^2 + y^2 - 5x - 9y - 16 \:=\: 0,\quad(6, \text{-}4)$

.
• May 24th 2009, 10:34 AM
Maths_noob
just checked it twice over, there is no typos (Doh)

thats the exact thing from my coursework...
• May 24th 2009, 11:04 AM
aidan
Quote:

Originally Posted by Maths_noob
Find the equation of the tangent to each circle at the point specified

1 Circle $x^2 + y^2 - 8x - 42y =$ 0, point (8, 5)

2 Circle $x^2 + y^2 - 5x - 9y - 16 = 0$ point(6, -4)

Any help would be great, don't even know what I should read up on to actually find out how to do it either... (Crying)

Thanks

In the above (1) is the isolated point (8,5) external to the circle, and you want the equation of a line from that point to a point tangent to the circle (of which there will be two lines tangent to the circle from the isolated point)?

Just to make sure that I understand the question?
• May 24th 2009, 11:19 AM
Maths_noob
well if i wanted it for

$x^2 + y^3 + 2x + 4y - 3 = 0$ at point (1, -4)

the answer should be along the lines of

$y - y1 = m2 (x - ^1)
y -2 = 2/3 (x - 0)
y = 2/3 x + 2$

therefore the y-intercept of this line is 2 and it passes through the give point (0,2)

I think that's the sort of thing the question is after..but i can't understand the steps taken lol
• May 24th 2009, 12:15 PM
HallsofIvy
Quote:

Originally Posted by Maths_noob
well if i wanted it for

$x^2 + y^3 + 2x + 4y - 3 = 0$ at point (1, -4)

Why did you change the problem. This is not a circle. Even if you meant $y^2$, not $y^3$ it would not be.
$x^2+ y^2+ 2x+ 4y- 3= x^2+ 2x+ 1+ y^2+ 4y+ 4- 3= 0$
$(x+ 1)^2+ (y+ 2)^2+ 2= 0$ which is never true. The "graph" is the empty set.

Quote:

the answer should be along the lines of

$y - y1 = m2 (x - ^1)
y -2 = 2/3 (x - 0)
y = 2/3 x + 2$

therefore the y-intercept of this line is 2 and it passes through the give point (0,2)

I think that's the sort of thing the question is after..but i can't understand the steps taken lol
For your original problem, $x^2 + y^2 - 8x - 42y = 0$, point (8, 5), $x^2- 8x+ 16+ y^2- 42y+ 441= 441$
$(x+ 4)^2+ (y- 21)^2= 21^2$ is a circle with center at (-2, 21) and radius 21.

Further, the distance from (-2, 21) to (5, 8) is less than 21 so this point is inside the circle. There is NO tangent to the circle through (5, 8)!

$x^2 + y^2 - 5x - 9y - 16 = x^2- 5x+ 25/4+ y^2- 9y+ 81/4- 25/4- 81/4- 16=0
$

$(x- 5/2)^2+ (y- 9/2)^2= 170/4$ is a circle with center at (5/2, 9/2) and radius $\sqrt{170}/2$ which is slightly larger than 6.5. The distance from (5/2, 9/2) to (6, -4) is $\sqrt{338}/2$, a little larger than 9.

Finally, a problem that can be done!

Geometrically, if you draw the lines from the given point tangent to the circle, the line from the given point to the center of the circle, and the line from the center of the circle to the point of tangency, you have a right triangle, the right angle being at the point of tangency. Your circle has radius $\sqrt{170}/2$ and the distance from (6, -4) to (5/2, 9/2) is $\sqrt{338}/2$ so you have a right triangle with hypotenuse of length $\sqrt{338}/2$ and one leg of length $\sqrt{170}/2$. By the Pythagorean theorem, the length of the other leg, and the distance from (8, 5) to the point of tangency, is $\sqrt{338- 170}/2=\sqrt{168/2}$. That means that we could find the points of tangency by striking a circle about (6, -4) with radius $\sqrt{168}/2$. That is, the points of tangency satisfy $(x-6)^2+ (y+4)^2= 42$ and $(x- 5/2)^2+ (y- 9/2)^2= 170/4$.
Solve those two equations for x and y to get the two points of tangency and, together with (6, -4), gives two points on each tangent line so you can find the equations.