Expert: Scott A Wilson - 2/10/2010

Question
QUESTION: Hi,  

This problem simply dazzles me:

The straight line with equation y = kx is a tangent to the circle x2 + y2 – 4x – 4y + 7 = 0.  Find the possible values of k giving your answers in the form a + b*sqrt7.

It's meant to be a GCE Advanced level Maths problem.  All I seem to be able to figure out from the geometry which relates vaguely to the hint is that the distance from the point of intersection from the tangent to origin is sqrt7.  Could you please solve it for me.  Thanks a million in advance.

ANSWER: Draw a right trianlge with one end at the origin, the other a the circle center,
and the other where the line is tangent to the circle.

Since it is tangent, this means that it is a right triangle.

Note the the hypoteneuse has length 2√2 and the far side has length 1.
Call the angle at the origin A.  The sin(A) of the angle is then 1/(2√2).

Note that this 45° is the angle of the hypoteneuse with the x-axis,
so 45° - A = B is the angle with the x-axis.

Now that you know the angle, the slope is the tan(B).


---------- FOLLOW-UP ----------

QUESTION: Thank you so much for this and the speed of response.  If I understand your solution, then k will be sinB/cosB.  This seems to exclude any chance of expressing the answer in the form a + b*sqrt7 as instructed and doesn't seem to agree with the suggestion that k has more than 1 value.  I do beg your pardon, but am I still missing something?

Answer
By the way, sin(x)/cos(x) = tan(x).  Yet now I see how to do it without sin() and cos().


We know that the length of the line that is tangent is the length of a side on a right triangle.
The far side is 1 and the hypoteneuse is 2√2, which makes that side √7.

Now with the point at which the line is tangent, the line is a hypoteneuse to a triangle
with the x-axis as its base.

Say the tangent point on the circle is (c,d).  This makes c² + d² = 7 in the triangle
that has the measure on the x-axis as c, a vertical line as d, and the line of length
√7 as the hypoteneuse.

It is also known that this point is on the circle and therefore (c-2)² + (d-2)² = 1.
Squareing the terms gives c² -4c + 4 + d² - 4d + 4 = 1.  Since we know c² + d² = 7,
put that in and get 7 - 4c + 4 + - 4d + 4 = 1.  Combining terms leads to 14 - 4c - 4d = 0.

From here, we can divide by 4 and get 3.5 - c - d = 0.  Add d to the other side,
flip the equation around, and get d = 3.5 - c.

Put this in the equation c² + d² = 7 and get c² + (3.5-c)² = 7.
That works out to c² + 12.25 - 7c + c² = 7.  Combining terms leads to 2c² - 7c + 5.25 = 0.

The quadratic equation gives us (7±√(49-42))/4 = (7±√7)/4 as c.

If c is taken as the term with a minus sign, I believe d would be the term with a plus sign.
That wold give us two points, on on each side of the circle.
They would be ((7-√7)/4,(7+√7)/4) and ((7+√7)/4 ,(7-√7)/4).

Checking both of these points out, it is found that they both lie on the circle
and both have a tangent line through the origin.

The equations of the lines would be y = ((7-√7)/4/(7+√7)/4)x and y = ((7+√7)/4/(7-√7)/4)x.
These simplify easily to y = ((7-√7)/(7+√7))x and y = ((7+√7)/(7-√7))x.

Multiplying the numerator and the denominator by the conjugate of the denominator yields
y = ((7-√7)²/42)x and y = ((7+√7)²/42)x.

That works out to be y = ((56-14√7)/42)x and y = ((56+14√7)/42)x.
Now that they're multiplied out, a 14 will cancel, giving
y = ((4-√7)/3)x and y = ((4+√7)/3)x.


If he wants it in a + b√7, it would be 4/3 + (1/3)√7 and  4/3 - (1/3)√7.
So a = 4/3 and b = ±1/3.


As an afterthought, to go back and look at the angle,
it seems like we would have tan(A) = (4+√7)/3 and tan(A) = (4-√7)/3
since this is the value of y when x is 1.