solve using any method

Question

x^2+y^2+4x=0

x^2+2y^2-4x+10y=0

Answer 1

x^2 +y^2 +4x = 0

y^2 = -x^2 -4x

x^2 +2y^2 -4x +10y = 0

x^2 +2(-x^2 -4x) -4x = -10y

-x^2 -8x -4x = -10y

x^2 + 12x = 10y

square both sides

x^4 +24x^3 +144x^2 = 100y^2 = 100(-x^2 -4x)

x^4 +24x^3 +244x^2 +400x = 0

x(x^3 +24x^2 +244x +400) = 0

one solution is x=0, y=0 (0,0)

the two original equations are a circle and ellipse which intersect potentially 4 times

but in this case, just twice, with 2 real solutions

(0,0) and (-2,-2)

(-2)^3 +24(-2)^2 + 244(-2) +400

=-8 + 96 -488 +400

= 0

x=-2 is the x coordinate of another intersection of the ellipse and circle

x^2 + y^2 -4x = 0

(-2)^2+ y^2 -4(-2) = 0

4 +y^2 = 8 = 0

y^2 = 8-4 =4

y = + or - sqr4 = 2 or -2

(-2,-2) is another solution

(-2)^2 +2(-2)^2 -4(-2) +10(-2)

=4 +8 +8 -20

=0

check (-2,2) as a 3rd possible 3rd solution. It satisfies the 1st original equation, but maybe not the 2n

(-2)^2 +2(-2)^2 -4(-2) +10(2)

4 + +8 +8 +20

=20+20= 40 which doesn't = 0

so there are just 2 solutions (0,0) and (-2,-2)

Answer 2

(x + 2)² + y² = 4; this is circle with (- 2, 0) center and r = 2

(x - 2)² + 2(y + 2.5)² = 16.5; (x - 2)²/16.5 + (y + 2.5)²/8.25 = 1; this is ellipse with center (2, - 2.5); a² = 16.5

and b² = 8.25