# Circle and a line that passes through it

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?

• What happens to the line $y=ax-3$ when $a$ gets really big, and does it pass through the circle? think in terms of the graph – user160738 Dec 16 '15 at 16:36
• The line $y=0$ does touch the circle - your circle has center $(1,1)$ and radius $1$. But $y=ax-3$ never actually becomes line $y=0$, it just gets closer and closer to that line. But still if the line is really steep, then it should pass through the circle in two different points. So try imagining $a$ getting smaller and smaller from there - as it gets smaller gradient of line gets smaller, and at some point it must be tangent to the circle. Is that clear? – user160738 Dec 16 '15 at 16:48
• well, substitute $y=ax-3$ into $(x-1)^{2}+(y-1)^{2}=1$ and solve the quadratic equation with respect to $a$ and see that for what range of $x$ you can find $a$ and then I think that you can find the possible range for $a$. – Albert Dec 16 '15 at 16:54
• It doesnt become $y=0$ neither $y=2$ neither $x=0$ neither $x=0$. If it does then it is a tangent to the circle. – User 2524 Dec 16 '15 at 16:55

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

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.

• How do i get the quadratic equation specifically? I'm a 10th grader so I dont quite understand the term "parametric". And also how do I get two unknowns with the circle equation: $(x-1)^2+(y-1)^2=1$ and the line equation: $y=ax-3$? Or does one of them only works with another equation? Thanks! – User 2524 Dec 16 '15 at 17:22
• If you take the equation of the circle, and replace the (only) occurrence of $y$ in that equation with the expression $ax-3,$ then you get an equation from which the variable $y$ has disappeared and you only need to solve it for the remaining unknown $x.$ Can you solve quadratic equations? – Justpassingby Dec 16 '15 at 17:24
• Of course :) What is the best way to understand linear-algebra? How can i improve? – User 2524 Dec 16 '15 at 17:40
• You don't need linear algebra for this one. Have you written down the quadratic equation in $x$? In particular, have you got an expression for its discriminant, which involves $a$? – Justpassingby Dec 16 '15 at 17:41
• Another guy passed by and solved it and in addition to your explaination, I guarantee I can solve similar problems. – User 2524 Dec 16 '15 at 17:47

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,

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