Question

Please help me I need help please​

Answer 1

Answer:

p = 4

Step-by-step 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