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InterviewSolution
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Find the equation of the circle passing through the point (7, 3) having radius 3 units and whose centre lies on the line `y = x-1` |
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Answer» Let equation of circle be `(x-h)^(2)+(y-k)^(2)=r^(2)…..(i)` `rArr (x-h)^(2)+(y-k)^(2)=9` Given that, centre (h,k) lies on the line y=x-1 i.e., k=h-1 Now, the circle passes through the point (7,3). `(7-h)^(2)+(3-k)^(3)=9` `rArr 49-14h+h^(2)+9-6k+k^(2)=9` `rArr h^(2)+k^(2)-14h-6k+49=0`.....(iii) On putting k=h-1 in Eq. (iii) we get `h^(2)+(h-1)^(2)-14h-6(h-1)+49=0` `rArr 2h^(2)-22h+56=0` `rArrh^(2)-11h+28=0` `rArr h^(2)-7h-4h+28=0` `rArrh(h-7)-4(h-7)=0` `rArr(h-7)(h-4)=0` `therefore h=4,7` When h=7, then k=7-1=6 `therefore` Centre(7,6) When h=4, then k=3 `therefore` Centre-(4,3) So, the equation of circle when centre (7,6) is `(x-7)^(2)+(y-6)^(2)=9` `rArr x^(2)-14x+49+y^(2)-12y+36=9` `rArr x^(2)+y^(2)-14x-12y+76=0` When centre (4,3), then the equation of the circle is `(x-4)^(2)+(y-3)^(2)=9` `therefore x^(2)-8x+16+y^(2)-6y+9=9` `rArr x^(2)+^(2)-8x-6y+16=0` |
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