Answer:
Tangent, not a tangent
Step-by-step explanation:
Given that a circle has equation as
[tex](x-5)^2+(y-1)^2=25[/tex]
We have to check whether the two lines are tangents to the circle
I line:
[tex]y−4=\frac{4}{3} (x−1)[/tex]
Substitute for y from straight line equation in the circle equation and check whether equal solutions are there. If equal solutions, then the line is tangent.
[tex](x-5)^2 + (\frac{4x}{3} +\frac{5}{3} )^2 =25\\x^2-10x+\frac{16x^2}{9} +\frac{25}{9} +\frac{40x}{9} =0\\25x^2-25x+25=0\\x=1,1[/tex]
Since equal roots are there this is a tangent
II line.
Substiutte for y to get
[tex](x-5)^2 +(\frac{3x}{4} -2)^2 =25\\x^2-10x +\frac{9x^2}{16} +4-3x=0\\25x^2-208x+64 =0\\[/tex]
Here discriminant not equals 0
Not equal roots
So cannot be a tangent.
