Question

4. Could a circle given by the equation (xx−5)2+(yy−1)2=25 have tangent lines given by the equations

Answer 1

Answer:

Tangent, not a tangent

Step-by-step explanation:

Given that a circle has equation as

$(x-5)^2+(y-1)^2=25$

We have to check whether the two lines are tangents to the circle

I line:

$y−4=\frac{4}{3} (x−1)$

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.

$(x-5)^2 + (\frac{4x}{3} +\frac{5}{3} )^2 =25\\x^2-10x+\frac{16x^2}{9} +\frac{25}{9} +\frac{40x}{9} =0\\25x^2-25x+25=0\\x=1,1$

Since equal roots are there this is a tangent

II line.

Substiutte for y to get

$(x-5)^2 +(\frac{3x}{4} -2)^2 =25\\x^2-10x +\frac{9x^2}{16} +4-3x=0\\25x^2-208x+64 =0$

Here discriminant not equals 0

Not equal roots

So cannot be a tangent.