find equations of the lines passing through the origin that are tangent to a circle with radius $2$ and center at point $(2,1)$ .

Full solution

Q. find equations of the lines passing through the origin that are tangent to a circle with radius

2

and center at point

(2,1)

Find Circle Equation: Find the equation of the circle. Equation: $(x - 2)^2 + (y - 1)^2 = 2^2$
Use Point and Line: Use the point $(0,0)$ and the general form of a line $y = mx$ to find the tangent lines.
Substitute and Simplify: Substitute $y = mx$ into the circle's equation. $(x - 2)^2 + (mx - 1)^2 = 4$
Expand and Simplify: Expand and simplify the equation. $(x - 2)^2 + (mx - 1)^2 = 4$ $x^2 - 4x + 4 + m^2x^2 - 2mx + 1 = 4$ $(1 + m^2)x^2 - (4 + 2m)x + 5 = 4$ $(1 + m^2)x^2 - (4 + 2m)x + 1 = 0$
Check Discriminant: For the line to be tangent, the discriminant of the quadratic equation must be zero. $b^2 - 4ac = 0$ $(-4 - 2m)^2 - 4(1 + m^2)(1) = 0$
Solve Discriminant: Solve the discriminant equation. $\newline$ $(4 + 2m)^2 - 4(1 + m^2) = 0$ $\newline$ $16 + 16m + 4m^2 - 4 - 4m^2 = 0$ $\newline$ $16m + 12 = 0$ $\newline$ $16m = -12$ $\newline$ $m = -\frac{3}{4}$
Find Line Equations: Find the equations of the lines using the slope $m$ . $y = \left( -\frac{3}{4} \right)x$
Check Other Solutions: Check for other possible values of $m$ . Since the discriminant equation is quadratic, there should be another solution. $(4 + 2m)^2 - 4(1 + m^2) = 0$ $16 + 16m + 4m^2 - 4 - 4m^2 = 0$ $16m + 12 = 0$ $16m = -12$ $m = -\frac{3}{4}$