Answer:
p = 4
Step-by-step explanation:
Given equation:
[tex]x^2+(p-3)y^2-4x+6y-16=0[/tex]
Standard equation of a circle:
[tex](x-a)^2+(y-b)^2=r^2[/tex]
(where [tex](a,b)[/tex] is the centre of the circle, and [tex]r[/tex] is the radius)
If you expand this equation, you will see that the coefficient of [tex]y^2[/tex] is always one.
Therefore, [tex]p-3=1[/tex]
[tex]\implies p=1+3=4[/tex]
Additional information
To rewrite the given equation in the standard form:
[tex]\implies x^2+y^2-4x+6y-16=0[/tex]
[tex]\implies x^2-4x+y^2+6y=16[/tex]
[tex]\implies (x-2)^2-4+(y+3)^2-9=16[/tex]
[tex]\implies (x-2)^2+(y+3)^2=16+4+9[/tex]
[tex]\implies (x-2)^2+(y+3)^2=29[/tex]
So this is a circle with centre (2, -3) and radius √29
