(1) In a sequence of circles C1, C2, C3, ....... Cn ; the centres lie along positive x-axis with abscissae forming an arithmetic sequence of first term unity and common difference 3. The radius of these circles are in geometric sequence with first term unity and common ratio 2. If the tangent lines with slope m1 and m2 of C3 are intersected at the centre of C5, then compute the value of 2010 \left|m_{1}m_{2} \right| . (2) The radii of the escribed circles of \Deltaï€ ABC are ra , rb and rc respectively. If ra + rb = 3R and rb + rc = 2R, then the smallest angle of the triangle is (A)\tan^{-1} (\sqrt{2}-1) (B)\frac{1}{2}\tan^{-1} (\sqrt{3}) (C)\frac{1}{2}\tan^{-1} (\sqrt{2}+1) (D)\tan^{-1} (2-\sqrt{3})
1for circle c3
center 7,0
radius 4
fir circle c5
certer 13,0
now we have to find the slope of the tangent of the circle c3
(x-7)2 + y2=42
which intersects at (13,0)
take a line(for the tangents) y=mx+c and satisfy two conditions:
1.it passes through (13,0)
2.touches the circle (x-7)2 + y2=42 i.e. is at a distance of 4 from (7,0)
u should get m2=4/5
1can u show me the method.....else can u just point out an error in my method!!!!!
1Well i started the same way as u
but instead of doing equations, I did as govind suggested and used trignometry
There is a triangle formed whose hypotenuse is 6, base is 4
So we can solve for tan theta, which is slope...
1thanks madhumith.....i found my error....a slight calculation glitch!!!!
1@madhumith...i liked ur trig. method really saves alot of time!!!!