find the points of intersection by solving the simultaneous equations x^2 + y^2 = 13 y=x+1

Question

x^2 + y^2= 13

y= x + 1

Answer 1

To find the intersection point of given equations,convert in to y = f(x) form.

x^2+y^2 = 13

y^2 = 13-x^2

Take square root to each side.

√y^2 =√13-x^2

y =√13-x^2 ----------> (1)

y = x+1 -----------> (2)

From (1) and (2) we also write x+1 =√(13-x^2)

Squring to each side.

(x+1)^2 =[√(13-x^2)]^2

x^2+1+2x = 13-x^2

Bring all the terms to left side.

x^2+1+2x-13+x^2 = 0

2x^2+2x-12 = 0

Divide by 2 to each side.

(2x^2+2x-12)/2 = 0/2

x^2+x-6 = 0

Factorize the above quadratic equation.

x^2+3x-2x-6 = 0

x(x+3)-2(x+3) = 0

(x-2)(x+3) = 0

x =2, x = -3

Intersection point to given simultaneous equations x^2+y^2 = 13, y = x+1 is (2, -3)

 

Comments

The points of intersection by solving the simultaneous equations x² + y² = 13 and y = x + 1 are (- 3, - 2) and (2, 3).

Answer 2

y = x + 1

Substitute y value in given equation

If x = -3 then y = x + 1==> y = -3 + 1 ==> y  = -2

If x = 2 then y = x + 1==> y = 2 + 1 ==> y  = 3