Find the tangent at the given point on the circle.

01 / Standard#Your turn

x^2+y^2+2x-6y-15=0,\quad T=(-4,-1)

Official paper · jm01-2026 · I.8 · PDF 2

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Complete squares to find the centre.

Hint 2

The tangent is perpendicular to the radius at T.

Worked solution

The centre-to-point vector is normal to the tangent.
$(x+1)^2+(y-3)^2=25,\ C=(-1,3),\ \overrightarrow{CT}=(-3,-4)$
Use the point-normal form.
$3(x+4)+4(y+1)=0\iff3x+4y+16=0$

D: 3x+4y+16=0.

Checks and common pitfalls: Use the centre, rather than the origin, to obtain the radius direction.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
The centre-to-point vector is normal to the tangent.
$(x+1)^2+(y-3)^2=25,\ C=(-1,3),\ \overrightarrow{CT}=(-3,-4)$
Use the point-normal form.
$3(x+4)+4(y+1)=0\iff3x+4y+16=0$

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

2026 JM01