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Find the equations of the circles passing through the point `(-4,3)`and touching the lines `x+y=2`and `x-y=2` |
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Answer» Correct Answer - `x^(2)+y^(2)+2(10pm3sqrt6)x+(55pm24sqrt6)=0` Let the equation of the required circle be `x^(2)+y^(2)+2gx+2fy+c=0" "...(i)` It passes through (-4, 3). `25-8g+6f+c=0" "...(ii)` Since, circle touches the line x + y - 2 = 0 and x - y - 2 = 0. `therefore |(-g-f-2)/(sqrt2)|=|(-g+f-2)/(sqrt2)|=sqrt(g^(2)+f^(2)-c)" "...(iii)` Now, ` |(-g-f-2)/(sqrt2)|=|(-g+f-2)/(sqrt2)|` `rArr -g-f-2=pm(-g+f-2)` `rArr -g-f-2=-g+f-2` or `rArr -g-f-2=g-f+2` `rArr f = 0 or g=-2` Case I when f = 0 From Eq. (iii) we get `|(-g-2)/(sqrt2)|=sqrt(g^(2)-c)` `rArr (g+2)^(2)= 2(g^(2)-c)` `rArr g^(2)-4g-4-2c=0" "...(iv)` On putting f = 0 in Eq. (ii) we get 25 = 8g + c = 0 ...(v) Eliminating c between Eqs. (iv) and (v),we get `g^(2)-20g+46=0` `rArr g = 10 pm3sqrt6 and c = 55pm24sqrt6` On substracting the value of g, f and c in Eq.(i) we get `x^(2)+y^(2)+2(10pm3sqrt6)x+(55pm24sqrt6)=0` Case II when g = - 2 From Eq. (iii), we get `rArr f^(2)=2(4+f^(2)-c)` `rArr f^(2)-2c + 8 = 0 " "...(vi)` On putting g = -2 in Eq. (ii), we get c = -6f - 41 ON substituting c in Eq.(vi), we get `f^(2)+12f+90=0` this equation gives imageinary value of F. Thus, there is no circle in this case. Hence, the required equations of the circles are `x^(2)+y^(2)+2(10pm3sqrt6)x+(55pm24sqrt6)=0` |
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