Question

Please help me I need help please​

Answer 1

p = 4

Explanation:

Given equation:

`x^2+(p-3)y^2-4x+6y-16=0`

Standard equation of a circle:

`(x-a)^2+(y-b)^2=r^2`

(where
`(a,b)` is the centre of the circle, and
`r` is the radius)

If you expand this equation, you will see that the coefficient of
`y^2` is always one.

Therefore,
`p-3=1`

`\implies p=1+3=4`

Additional information

To rewrite the given equation in the standard form:

`\implies x^2+y^2-4x+6y-16=0`

`\implies x^2-4x+y^2+6y=16`

`\implies (x-2)^2-4+(y+3)^2-9=16`

`\implies (x-2)^2+(y+3)^2=16+4+9`

`\implies (x-2)^2+(y+3)^2=29`

So this is a circle with centre (2, -3) and radius √29