Circle and a line that passes through it

Question

Given a line with equation: $y=ax-3$ that passes through a circle with equation $(x-1)^2+(y-1)^2= 1$. Find the range of values of $a$.

I tried graphing and got: $0<x<2$ and $0<y<2$.

I also tried finding $a$ by substituting $x$ and $y$ into $y=ax-3$ which really confuses me.

Could you help me in solving this problem?

Answer 1

we substitute
$y=ax-3$
in
$(x-1)^2+(y-1)^2= 1$

$(x-1)^2+(ax-3-1)^2= 1$
$(x-1)^2+(ax-4)^2= 1$
$x^2-2x+1+a^2x^2-8ax+16=1$
we have this second degree equation

$x^2(1+a^2)+x(-2-8a)+16=0$
with

$\Delta =(-2-8a)^2-4\cdot(1+a^2)\cdot 16$
if
$\Delta <0$
the equation have no solution and the line not intersect the circle
if
$\Delta =0$
the equation have one solution and the line is tangent to the circle
if
$\Delta >0$
the equation have two solutions and the line intersect the circle in 2 points

$\Delta =4+32a+64a^2-64-64a^2$
$\Delta =32a-60$

so if
$32a-60>0$
$a>\frac{15}{8}$
the line intersect the circle in 2 points

and if
$32a-60=0$
$a=\frac{15}{8}$
the line is tangent to the circle

Answer 2

The lines with equations $y=ax-3$ all have in common that they intersect the $Y$ axis at $(0;-3).$ Which of those lines intersect the circle with centre $(1;1)$ and radius 1?

The equation of the circle and the (parametric) equation of the line form a system of 2 simultaneous equations with 2 unknowns. One of the equations is nonlinear but the system is easy enough to solve.

At one stage in the solution you will have to use the roots of a quadratic equation. The condition on $a$ that you are looking for, is for the discriminant of that quadratic equation to be nonnegative.

Answer 3

Rewrite the system of equations by substituting $y$, giving

$$(x-1)^2+(ax-3-1)^2-1=0=(a^2+1)x^2-(8a+2)x+16.$$

The latter equation has roots when its discriminant is non-negative,

$$(4a+1)^2-16(a^2+1)=8a-15\ge0.$$