



 
Hi Sandra, First you need to find the equations of each of the circles then you need to solve the system of equations. I'll use a similar example where the equations are already found: (x  2)^{2} + (y  5)^{2} = 25 (x  6)^{2} + (y  13)^{2} = 65 First expand the binomials and collect like terms to one side of the equation: x^{2}4x+4 + y^{2} 10y+25=25 ⇒x^{2}4x+y^{2} 10y+4=0 (Eq. 1) x^{2}12x+36+y^{2}26y+169=65 ⇒ x^{2}12x+y^{2}26y+140=0 (Eq. 2) Since both equations now equal 0, we can set them equal to each other: x^{2}4x+y^{2} 10y+4=x^{2}12x+y^{2}26y+140 Next collect like terms 8x+16y136=0 then solve for a variable 8x=13616y x=172y (Eq. 3) Substitute into one of the expanded equations (I choose eq. 1) x^{2}4x+y^{2} 10y+4=0 ⇒ (172y)^{2}4(172y)+y^{2} 10y+4=0 ⇒ 28968y+4y^{2}68+8y+y^{2}10y+4=0 and then collect like terms 5y^{2}70y+225=0 We can factor this binomial and then find the roots 5(y^{2}14y+45)=0 5(y9)(y5)=0 ⇒ y=9 and y=5 these are the yvalues of our intersection points. to find the corresponding xvalues, substitute into Eq.3 x=172(9)=1 x=172(5)=7 So our intersection points are (1,9) & (7,5) Hope this helps, Janice  


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