A circle has centre (1,−4) and radius 5. Find its equation and the length of a tangent drawn from the point (6,2) to this circle.

Step-by-step solution

1. Write the general equation of a circle: $(x-1{)}^2+(y+4{)}^2=5^2$.
2. Expand the radius part: $5^2=25$, so the equation becomes $(x-1{)}^2+(y+4{)}^2=25$.
3. Compute the distance from the external point (6,2) to the centre (1,−4): $\sqrt{(6-1{)}^2+(2+4{)}^2}=\sqrt{25+36}=\sqrt{61}$.
4. Use the tangent length formula: $t=\sqrt{d^2-r^2}$ where $d$ is the distance to the centre and $r$ is the radius.
5. Substitute $d^2=61$ and $r^2=25$: $t=\sqrt{61-25}=\sqrt{36}=6$.
6. Thus the length of the tangent is 6.

Concept tested: Circle equation & tangent length

Published 16 October 2026 · How this solution was prepared and checked