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1. Hi,

1. Show that the line with eqn. 2x-3y+26=0 is a tangent to the circle with the eqn. x^2 + y^2 - 4x + 6y - 104 = 0. The point of contact is T, find the co-ordinates of the point on the circle which is diametrically opposite to T.

2. Line with eqn. y=mx+c is tangent to the circle with eqn. x^2+y^2-6x-6y+17=0.
Find the possible values of m.

3. Prove that the circle with eqn. x^2+y^2-2ax-2by+b^2=0 touches the y-axis. Hence, or otherwise, find the eqns. of the 2 circles which pass through the points (1,2) and (2,3) and which touch the y-axis.
2. find y in terms of x in the linear eq, then sub that in to the circle, to find Y, then sub y in to the linear to find T.
since the linears gradient is 2/3, and is a tangent, so, the radius that it meets has a gradient of -3/2
use y-y1=m(x-x1) to find eq of the radius, then, set that equation equal to the circle, which should find T, and the other point, diametrically opposite to T

this is how i would attempt it, please bear in mind, im a C2 student, who, for an hour a week, reads P3
3. anyone?
4. i tried subbing in x=1.5y-13 but it doesnt seem to work
5. (Original post by matt.forster)
find y in terms of x in the linear eq, then sub that in to the circle, to find Y, then sub y in to the linear to find T.
since the linears gradient is 2/3, and is a tangent, so, the radius that it meets has a gradient of -3/2
use y-y1=m(x-x1) to find eq of the radius, then, set that equation equal to the circle, which should find T, and the other point, diametrically opposite to T

this is how i would attempt it, please bear in mind, im a C2 student, who, for an hour a week, reads P3
not bad just a few tweaks...

find y in terms of x in the linear equation. rewrite the circle eq into proper circle form, ie (x-a)^2 + (y-b)^2 = (r^2). then sub the expression for y in terms of x and solve to show only one solution. put this value of x back into the linear equation to find the value of y.

2x-3y+26=0
y = (2x + 26)/3

x^2 + y^2 - 4x + 6y - 104 = 0
(x-2)^2 + (y+3)^2 = 117
(x-2)^2 + [(2x + 26 + 9)/3]^2 = 117
(x-2)^2 + (1/9)[2x + 35]^2 = 117
(x^2) - 4x + 4 + (1/9)[4(x^2) + 140x + 1225] = 117
(1/9)[(9x^2) - 36x + 36 + 4(x^2) + 140x + 1225] = 117
(9x^2) - 36x + 36 + 4(x^2) + 140x + 1225 = 1053
13(x^2) + 104x + 1261 = 1053
13(x^2) + 104x + 208 = 0
(x^2) + 8x + 16 = 0

determinant = b^2 - 4ac
= (8^2) - 4(1)(16)
= 64 - 64
= 0

determinant = 0 => one repeated real root => tangent as only one common point...

(x^2) + 8x + 16 = 0
(x+4)(x+4) = 0
x = -4

y = (2x + 26)/3
y = -8 + 26)/3
y = 6
6. for the first one ive got it now, you have to sub in y=2/3x+26/3

x^2 + (2/3x+26/3)^2 - 4x + 6(2/3x+26/3) - 104

13/9x^2 + 104/9x + 208/9 = 0

sqr root of b^2-4ac = 104^2 - 4 x 13 x 208 = 0

this means its a tangent
7. can you please answer 3, and i have another question, if your given the eqn of a circle, (x-a)^2 + (y-b)^2 = r^2. and the line OT is a tangent to the circle.
how do you find an eqn. of the line OT.
8. (Original post by manps)
can you please answer 3, and i have another question, if your given the eqn of a circle, (x-a)^2 + (y-b)^2 = r^2. and the line OT is a tangent to the circle.
how do you find an eqn. of the line OT.
Use pythagoras as the tangent makes a right angle with a radius.
9. 3.
If the circle touches the y-axis, then there's only 1 point of intersection between it and the line x=0. That is:
x² + y² - 2ax - 2by + b² = 0
y² - 2by + b² = 0

D = B² - 4AC = (2b)² - 4(1)(b²) = 0, hence there is only one point of contact. QED.

Now let's plug (1,2) and (2,3) into the equation of the circle. Upon simplifcation we get:
b² - 4b - 2a + 5 = 0
b² - 6b - 4a + 13 = 0

Eliminating a:
b² - 2b - 3 = 0
(b-3)(b+1) = 0
b=3 or b=-1

Now find the corresponding values of a, and then you can use the pair of values of a & b to find the two equations.

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