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A is a point (a, b) in the first quadrant. If the two circIes which passes through A and touches the coordinate axes cut at right angles then :A. `a^(2)-6ab+b^(2)=0`B. `a^(2)+2ab-b^(2)=0`C. `a^(2)-4ab+b^(2)=0`D. `a^(2)-8ab+b^(2)=0` |
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Answer» Correct Answer - 3 Let the equation of the two circles be `(x-r)^(2)+(y-r)^(2)=r^(2)` i.e.., `x^(2)+y^(2)-2rx-2ry+r^(2)=0`, where `r=r_(1)r_(2)` The condition of orthogonality gives `2r_(1)r_(2)+2r_(1)r_(2)=r_(1)^(2)+r_(2)^(2)` or `4r_(1)r_(2)=r_(1)^(2)=r_(2)^(2)` (1) Circle passes through `(a,b)`. Therefore, `a^(2)+b^(2)-2ra-2rb+r^(2)=0` i.e., `r^(2)-2r(a+b)+a^(2)+b^(2)=0` `:. r_(2) +r_(2)=2(a+b)` and `r_(1)r_(2)=a^(2)+b^(2)` `:. 4(a^(2)+b^(2))=4(a+b)^(2)-2(a^(2)+b^(2))` [From (1)] i.e., `a^(2)-4ab+b^(2)=0` |
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