If the circle $$x2+y2+2x+3y+1=0$$ cuts $$x2+y2+4x+3y+2=0$$ in A and B, then the equation of the circle on AB as diameter, is

Question

If the circle $$x2+y2+2x+3y+1=0$$ cuts $$x2+y2+4x+3y+2=0$$ in A and B, then the equation of the circle on AB as diameter, is

Infinity Learn · Accepted Answer

The equation of the common chord AB of the two circles is       $$2x+1=0$$                            [Using $$S1-S2=0$$]The equation of the required circle is     $$x2+y2+2x+3y+1+$$λ$$(2x+1)=0$$                                                 [Using : $$S1+$$λ$$(S2-S1)=0$$]⇒ $$x2+y2+2x($$λ$$+1)+3y+$$λ$$+1=0$$.Since AB is a diameter of this circle. Therefore, centre of this circle lies on AB.∴ −$$2$$λ−$$2+1=0$$⇒λ$$=$$−$$1/2So$$, the equation of the required circle is    $$x2+y2+x+3y+1/2=0or$$, $$2x2+2y2+2x+6y+1=0$$

If the circle $x^{2} + y^{2} + 2 x + 3 y + 1 = 0$ cuts $x^{2} + y^{2} + 4 x + 3 y + 2 = 0$ in $A$ and $B,$ then the equation of the circle on $A B$ as diameter, is

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If the circle x2+y2+2x+3y+1=0 cuts x2+y2+4x+3y+2=0 in A and B, then the equation of the circle on AB as diameter, is

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If the circle $x^{2} + y^{2} + 2 x + 3 y + 1 = 0$ cuts $x^{2} + y^{2} + 4 x + 3 y + 2 = 0$ in $A$ and $B,$ then the equation of the circle on $A B$ as diameter, is