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Question Number 199392 by Calculusboy last updated on 03/Nov/23

Answered by mr W last updated on 03/Nov/23

(x−3)^2 +(y+4)^2 =53 (x+2)^2 +(y−1)^2 =13 eqn. of intersection line: (x−3)^2 −(x+2)^2 +(y+4)^2 −(y−1)^2 =53−13 −5(2x−1)+5(2y+3)=40 −x+y=2 ⇒y=x+2 intersection points: (x−3)^2 +(x+2+4)^2 =53 x^2 +3x−4=0 (x−1)(x+4)=0 ⇒x_1 =−4, x_2 =1 ⇒y_1 =−2, y_2 =3 i.e. intersection at (−4,−2) and (1,3)

(x−3)2+(y+4)2=53(x+2)2+(y−1)2=13eqn.ofintersectionline:(x−3)2−(x+2)2+(y+4)2−(y−1)2=53−13−5(2x−1)+5(2y+3)=40−x+y=2⇒y=x+2intersectionpoints:(x−3)2+(x+2+4)2=53x2+3x−4=0(x−1)(x+4)=0⇒x1=−4,x2=1⇒y1=−2,y2=3i.e.intersectionat(−4,−2)and(1,3)

Commented by Calculusboy last updated on 03/Nov/23

thanks sir

thankssir

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