The variable line $y=3x+b$ with gradient $3$ meets the circle $x^2+y^2=16$ at A and B. Find the locus of the midpoint M of AB.

Question

So far I have tried to solve the equation simultaneously $10x^2+6bx+(b^2-16)$ and I have found $x=\frac{-3b}{10}$ using the sum of roots but I don't know what to do after.

Thanks in advance :)

The midpoint of this secant $AB$ will lie on the radius that is perpendicular to the line. — Doug M, Apr 20 '17 at 02:09
To get the y-value, do the same thing, just start again and solve for y. — quasi, Apr 20 '17 at 02:09
But the hint by Doug M is a good one. Simpler (no quadratics). — quasi, Apr 20 '17 at 02:11

score 0 · Answer 1 · answered Apr 20 '17 at 02:43

When a line intersects a circle at points A and B the line segment AB is in fact a chord of that circle. Now you are varying the line. All these varations are parallel line, as all have slope 3. So you have to trace the midpoints of parallel chords. The shape traced out by mid-points of chords, by means of the perfect symmetry of the circle, is the same irrespective of the slope of the family of these chords.

Can you visualize this? Just draw picture. You can guess the locus. You can bring c0-ordinates after this stage.

The variable line $y=3x+b$ with gradient $3$ meets the circle $x^2+y^2=16$ at A and B. Find the locus of the midpoint M of AB.

1 Answers1