Question 6. A point $P(x,y)$ is on the circle

Key concept: This question concerns a point on a circle and a tangent line at that point. The key algebraic idea is to substitute the line into the circle and use the repeated-root condition.

Step by step

The circle is

$x^2+y^2-6x+4y-12=0.$

If a line is tangent to the circle at a point $P$, then after substitution the intersection equation becomes a quadratic with a repeated root.

Step 1: Rewrite the circle if needed

You can complete the square:

$x^2-6x+y^2+4y=12$

$(x-3)^2+(y+2)^2=25.$

So the circle has center $(3,-2)$ and radius $5$.

Step 2: Use the tangent condition

For a tangent through a point on the circle, the substituted equation must satisfy

$b^2-4ac=0.$

Step 3: Solve for the unknown parameter

The exact values depend on the intended line equation. In the pasted working, several coefficients change between lines, so the safest approach is to rebuild the algebra from the original line before solving.

Geometric check

Any valid answer should also make sense geometrically: the tangent at $P$ must be perpendicular to the radius from the center $(3,-2)$ to $P$.

Pitfall alert

A common mistake is to rely on an intermediate expansion that contains inconsistent coefficients. Another error is forgetting that the tangent at a point is perpendicular to the radius through that point, which can be used as a quick check on the algebra.

Try different conditions

If the point $P$ is given explicitly, the tangent line can be found using the gradient of the radius or by the point-to-line normal form. If instead only the slope of the tangent is given, use the circle equation to find the matching point(s) by discriminant or perpendicularity.

Further reading

circle center, tangent line, complete the square