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C2 circle problems

Im having trouble with these two problems.....

Prove that the circle (x+4)^2 + (y-5)^2 = 8^2 lies completely inside the
circle x^2 +y^2 +8x - 10y = 59




The circle C has equation

x^2 +y^2 - 10x + 4y +20 = 0

Find the length of the tangent to C from the point (-4,4)
Original post by chowderspin
Im having trouble with these two problems.....

Prove that the circle (x+4)^2 + (y-5)^2 = 8^2 lies completely inside the
circle x^2 +y^2 +8x - 10y = 59


Complete the squares on the second equation to get it into the same form as the first and something should strike you.



The circle C has equation

x^2 +y^2 - 10x + 4y +20 = 0

Find the length of the tangent to C from the point (-4,4)


The line joining the point to the centre of the circle, and the tangent line, together with a radius forms a right angle triangle. So work out the length of the two others and use pythagoras.
Reply 2
Avatar for IanBlue
IanBlue
I still couldn't get it...
CA(tangent point) = 3 [ Because its the radius ]
CP i have as root117 (3root13) [C(5,-2)]
Then Finally for A(tangent point)to P : root(3^2 + 117) = 3root14 =root126=11.2 [NOT 6ROOT3 OR 10.4 WHICH THE BOOK READS.]
Reply 3
Avatar for TenOfThem
TenOfThem
18
Original post by IanBlue
I still couldn't get it...
CA(tangent point) = 3 [ Because its the radius ]
CP i have as root117 (3root13) [C(5,-2)]
Then Finally for A(tangent point)to P : root(3^2 + 117) = 3root14 =root126=11.2 [NOT 6ROOT3 OR 10.4 WHICH THE BOOK READS.]


117\sqrt{117} is the hypotenuse!

Did you draw a diagram?
Reply 4
Avatar for IanBlue
IanBlue
Original post by TenOfThem
117\sqrt{117} is the hypotenuse!

Did you draw a diagram?




AHHH, i did. but it was a plotter triangle on a slant - thanks, i have the correct answer now

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