Tangents drawn from the point (4, 3) to the circle `x^(2)+y^(2)-2x-4y=0` are inclined at an angle

Options

A. `pi//6`
B. `pi//4`
C. `pi//3`
D. `pi//2`

Answer

D

Explanation

Using `SS_(1)=T^(2)`, the combined equation of the tangents drawn from (4, 3) to the circle `x^(2)+y^(2)-2x-4y=0`, is
`(x^(2)+y^(2)-2x-4y)(16+9-8-12)`
`{4x+2y-(x+y)-2(y+3)}^(2)`
`rArr 5 (x^(2)+y^(2)-2x-4y)=(3x+y-10)^(2)`
`rArr 4x^(2)-4y^(2)-50x+6xy+100=0 " " ...(i)`
In this equation , we have
Coeff. of `x^(2)+` Coeff. of `y^(2)=0`.
Therefore, lines given by (i) are at right angle to each other.
`ul("ALITER")` The equation of the given circle and its director circle are
`(x-1)^(2)+(y-2)^(2)=5` and `(x-1)^(2)+(y-2)^(2)=10 " " (i)`
Clearly, (4, 3) lies on (i). So, required angle is `pi//2`.