InterviewSolution
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InterviewSolution
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Find the equation of the circle which circumscribes the triangle formedby the line: ` y=x+2,3y=4x a n d 2y=3x` |
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Answer» Given equations of line are y=x+2…(i) 3y=4x….(ii) 2y=3x ….(iii) From Eqs. (i) and (ii) `(4x)/3=x+2` `rArr4x=3x+6` `rArr x=6` On putting x=6 in Eq.(i) we get y=8 `therefore` Point,A=(6,8) From Eqs. (i) and (iii), `(3x)/2=x+2` `rArr3x=2x+4rArrx=4` When x=4 then y=6 `therefore` Point, B=(4,6) From Eqs. (ii) and (iii) `x_(1)=0_(1),y=0` Now, C=(0,0) Let the equation of circle is `x^(2)+y^(2)+2gxn+2fy+c=0` since,the points A(6,8),B(4,6) and C(0,0) lie on this circle 36+64+12g+16f+c=0 `rArr 12g+16f+c=-100....(iv)` and 16+36+8g+12f+c=0...(v) `rArr 8g+12f+c=-52`....(vi) `rArr c=0` From Eqs. (iv) , (v) and (vi), 12g+16f=-100 `rArr3g+4f+25=0` `rArr2g+3f+13=0` `rArrg/(+52-75)=f/(50-39)=1/(9-8)` `rArrg/(-23)=f/11=1/1` `rArrg=-23,f=11` So, the equation of the circle is `x^(2)+y^(2)-46x+22y+0=0` `rArrx^(2)+y^(2)-46x+22y=0` |
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