Equidistant points and circle radius

See if you draw in the approximate position of P that meets the criteria. This should help you think about the problem more. Tell us what you think.
 
Let [imath](x,y)[/imath] be the coordinate of Point P

Given:
"Point P is also the same distance from point A as it is from point B" [imath]\Rightarrow \overline{\rm AP} = \overline{\rm BP}\\ \Rightarrow \sqrt{(1-x)^2+(3-y)^2}=\sqrt{(9-x)^2+(7-y)^2}---Equation(1)[/imath]
Task 1: Simplify Equation 1. It should look something like y=f(x), meaning y is on one side, and x will be on the other side of the equation.

"Points P and Q both lie on the same circle with centre C"[imath]\Rightarrow \overline{\rm PC} = \overline{\rm QC}[/imath] because both lengths are radii of the circle C.

Task 2: Can you find [imath]\overline{\rm QC}[/imath] ?
[imath]\overline{\rm QC}= \overline{\rm PC}=\sqrt{(12-x)^2+(6-y)^2}---Equation(2)[/imath]
Task 3: Now, you have 2 equations and 2 unknowns x,y. Solve for x and y.
 
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