If you're given a equation of a circle: x^2 + y^2 - 2x + 2y - 23 = 0 and an external point E(0.8),
The center of the circle is C(1,-1) and the radius is r= 5
The only given is the equation of the circle and the external point of the tangent line...
Find the POINTs OF TANGENCY and the two equation of the tangent lines.
NOTE: E(0,8) is the point of intersection of the two tangent lines, and can be found outside the circle..
Help please....
The center of the circle is C(1,-1) and the radius is r= 5
The only given is the equation of the circle and the external point of the tangent line...
Find the POINTs OF TANGENCY and the two equation of the tangent lines.
NOTE: E(0,8) is the point of intersection of the two tangent lines, and can be found outside the circle..
Help please....
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We can find the distance from the center of the circle to the external point:
d^2 = (8 - (-1))^2 + (0 - 1)^2 --> d^2 = 81 + 1 --> d = sqrt(82)
Now, wherever the points of tangency are on the circle, let's say at (x1, y1) and (x2, y2), we know that they are at points on the circle where the radius and tangent lines meet at right angles.
So: Each of those points, along w/ the external point and center of the circle, form right triangles.
Therefore: Using Pythagorean Theorem, we have
(sqrt(82))^2 - 25 = (x1 - 0)^2 + (y1 - 8)^2
--> 82 - 25 = x1^2 + y1^2 - 16y1 + 64 --> x1^2 + y1^2 = 16y1 - 7
But: Because (x1, y1) is on the circle, we also have
x1^2 + y1^2 - 2x1 + 2y1 - 23 = 0 Doing a substitution for x1^2 + y1^2:
(16y1 - 7) - 2x1 + 2y1 - 23 = 0 --> 18y1 - 2x1 - 30 = 0 --> x1 = 9y1 - 15
Putting this into our 1st (Pythagorean Theorem result) equation, we get:
(9y1 - 15)^2 + y1^2 = 16y1 - 7 --> 81y1^2 - 270y1 + 225 + y1^2 = 16y1 - 7
--> 82y1^2 - 286y1 + 232 = 0 --> 41y1^2 - 143y1 + 116 = 0
Solving this for y1 requires the quadratic formula (I'll leave that to you...), giving us two possible solutions: y1 = 1.28 or 2.20 (approximately)
These two solutions yield x1 = 9*1.28 - 15, or - 3.48 AND x1 = 9*2.20 - 15, or 4.8
SO: The tangent lines intersect the circle at (- 3.48, 1.28) and (4.8, 2.20) (approximately)
(NOTE: I drew a fairly accurate sketch, and these point coordinate values looked reasonable...)
That was the hard part. Now that you have the points of intersection, along with the external point of the tangent lines, you should be able to find the two equations of the tangent lines.
Good luck!
d^2 = (8 - (-1))^2 + (0 - 1)^2 --> d^2 = 81 + 1 --> d = sqrt(82)
Now, wherever the points of tangency are on the circle, let's say at (x1, y1) and (x2, y2), we know that they are at points on the circle where the radius and tangent lines meet at right angles.
So: Each of those points, along w/ the external point and center of the circle, form right triangles.
Therefore: Using Pythagorean Theorem, we have
(sqrt(82))^2 - 25 = (x1 - 0)^2 + (y1 - 8)^2
--> 82 - 25 = x1^2 + y1^2 - 16y1 + 64 --> x1^2 + y1^2 = 16y1 - 7
But: Because (x1, y1) is on the circle, we also have
x1^2 + y1^2 - 2x1 + 2y1 - 23 = 0 Doing a substitution for x1^2 + y1^2:
(16y1 - 7) - 2x1 + 2y1 - 23 = 0 --> 18y1 - 2x1 - 30 = 0 --> x1 = 9y1 - 15
Putting this into our 1st (Pythagorean Theorem result) equation, we get:
(9y1 - 15)^2 + y1^2 = 16y1 - 7 --> 81y1^2 - 270y1 + 225 + y1^2 = 16y1 - 7
--> 82y1^2 - 286y1 + 232 = 0 --> 41y1^2 - 143y1 + 116 = 0
Solving this for y1 requires the quadratic formula (I'll leave that to you...), giving us two possible solutions: y1 = 1.28 or 2.20 (approximately)
These two solutions yield x1 = 9*1.28 - 15, or - 3.48 AND x1 = 9*2.20 - 15, or 4.8
SO: The tangent lines intersect the circle at (- 3.48, 1.28) and (4.8, 2.20) (approximately)
(NOTE: I drew a fairly accurate sketch, and these point coordinate values looked reasonable...)
That was the hard part. Now that you have the points of intersection, along with the external point of the tangent lines, you should be able to find the two equations of the tangent lines.
Good luck!