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The circles `ax^(2)+ay^(2)+2g_(1)x+2f_(1)y+c_(1)=0" and "bx^(2)+by^(2)+2g_(2)x+2f_(2)y+c_(2)=0` `(ane0and bne0)` cut orthogonally, ifA. `g_(1)g_(2)+f_(1)f_(2)=ac_(1)+bc_(2)`B. `2(g_(1)g_(2)+f_(1)f_(2))=bc_(1)+ac_(2)`C. `bg_(1)g_(2)+af_(1)f_(2)=bc_(1)+ac_(2)`D. `g_(1)g_(2)+af_(1)f_(2)=c_(1)+c_(2)` |
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Answer» Correct Answer - 2 Given circles can be written as `x^(2)+y^(2)+(2g_(1))/(a)x+(2f_(1))/(a)y+(c_(1))/(a)=0` `x^(2)+y^(2)+(2g_(2))/(b)x+(2f_(2))/(b)y+(c_(2))/(b)=0` centre of circles are `C_(1)=((-g_(1))/(a),(-f_(1))/(a)),` `C_(2)=((-g_(2))/(a),(-f_(2))/(a)),` If 2 circules cut orthogonally `So,2((-g_(1))/(a))((-g_(2))/(b))+2((-f_(1))/(a))((-f_(2))/(b))=(c_(1))/(a)+(C_(2))/(b)` `rArr2(g_(1)g_(2)+f_(1)f_(2))=bc_(1)+ac_(2)` |
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