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If the circles x^2 + y^2 + 2ax + b = 0 a...

If the circles `x^2 + y^2 + 2ax + b = 0` and `x^2+ y^2+ 2cx + b = 0` touch each other `(a!=c)`

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The correct Answer is:
`a to s; b to r; c to q; d to p`
a. `{cos(2A + theta + cos (2B + theta)} = 2cos (A-B) cos (A+B+theta)`
Maximum value is ` 2cos(A-B)` when
`cos (A+B+theta)=1`
b. `{cos 2A + cos 2B} = 2cos (A+B) cos(A-B)`
Maximum value is `2cos(A-B)` when
`cos (A+B)=1`
c. For `y = secx , x in (0, pi//2)`, tangent drawn to it at any point lies completerly below the graph of `y = secx`, thus,
`" "(sec2A + sec2B)/(2) ge 2 sec(A+B)`
or `sec2A + sec2b ge 2 sec(A+B)`
Hence, the minimum value is `2 sec (A+B)`
d. `sqrt({tan theta + cot theta - 2 cos 2 (A +B)})`
`" " =sqrt((sqrt(tan theta) - sqrt(cot theta))^(2)+ 2-2 cos 2 (A+B)))`
Minimum value occurs when `sqrt(tan theta)= sqrt(cot theta and `
minimum value is `sqrt(4 sin ^(2) (A+B)) = 2 sin (A+B)`
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