Show that the line `3x - 4y -c = 0` will meet the circle having centre at (2, 4) and the radius 5 in real and distinct points if `-35 lt c lt 15`.

Explanation

Given line is `3x - 4y -c = 0" "…(i) `
Centre of given circle is (2, 4) and its radius is 5, therefore its equation will be `(x-2)^(2) + (y-4)^(2) = 5^(2)`
or `x^(2) + y^(2) - 4x - 8y - 5 = 0 " "…(ii)`
From (1), `y = (1)/(4) (3x - c)` Putting the value of y in (ii), we get
` x^(2) + (1)/(16) (3x -c)^(2) - 4x - 8*(1)/(4) (3x-c)-5 =0`
or `16x^(2) + 9x^(2) + c^(2) -6cx - 64x - 96x + 32c - 80 = 0`
or `25x^(2) - 2(80 + 3c) x + c^(2) + 32c - 80 = 0 " "...(iii)`
Line (i) will meet the circle (ii) in real and distinct points if discriminant of equation (iii) `lt 0`
i.el., if `4(80 + 3c)^(2) - 100(c^(2) + 32c - 80) gt 0`
or `(80 + 3c)^(2) -25(c^(2) + 32xc - 80) gt 0`
or `6400 + 9c^(2) + 480c - 25c^(2) - 800c + 2000 lt 0`
`rArr -16c^(2) - 320c + 8400 lt 0`
`rArr c^(2) - 20c - 525 lt 0`
`(c + 35)(c-15) lt =0`
`rArr -35 lt c lt 15`.