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Tangent to the curve `y=x^2+6` at a point `(1,7)` touches the circle `x^2+y^2+16x+12y+c=0 `at a point `Q`, then the coordinates of `Q` are (A) `(-6,-11)` (B) `(-9,-13)` (C) `(-10,-15)` (D) `(-6,-7)`A. 195B. 185C. 85D. 95 |
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Answer» Correct Answer - D Key idea Equation of tangent to the curve `x^(2)= 4ay "at"(x_(1), y_(1)) is "xx"_(1)=4a((y+y_(1))/(2))` Tangent to the curve `x^(2)=y-6 "at" (1,7) " is "` `x = (y+7)/(2)-6` `rArr 2x-y+5=0" "(i)` Equation of circle is `x^(2)+y^(2)+16x + 12y+C=0 " Centre"(-8,-6)` `r=sqrt(8^(2)+6^(2)-c)=sqrt(100-c)` Since, line 2x - y + 5 = 0 also touches the circle. `therefore sqrt(100-c)=|(2(-8)-(-6)+5)/(sqrt(2^(2)+1^(2)))|` `rArr sqrt(100-c)=|(-16+6+5)/(sqrt5)|` `rArr sqrt(100-c)=|-sqrt5|` `rArr 100-c=5` `rArr c= 95` |
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