The second distinct tangent from the origin to the circle is y=mx. Find m.

01 / Standard#Your turn

(x-4)^2+(y-4)^2=16/5,\quad L_2:y=mx

Official paper · jm01-2021 · II.2(b) · PDF 4

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Its perpendicular distance from M is also 4/√5.

Hint 2

Reject the slope corresponding to the already given tangent.

Worked solution

m=1/2.

Checks and common pitfalls: Both algebraic roots are tangent slopes, but only one is the requested second line.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Equate the squared centre-to-line distances.
$\frac{(4m-4)^2}{m^2+1}=16/5$
Solve and distinguish the two tangents.
$5(m-1)^2=m^2+1\iff(2m-1)(m-2)=0$
The root m=2 is L₁; the distinct second line has the other root.
$m=1/2$

Think first. Reveal a hint when the class is ready.

2021 JM01