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Tangent to the curve y = x2 + 6 at a point (1, 7) touches the circle x2 + y2 + 16x + 12y + c = 0 at a point Q. Then the coordinates of Q are (2005S)
- a)(–6, –11)
- b)(–9, –13)
- c)(–10, –15)
- d)(–6, –7)
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Tangent to the curve y = x2 + 6 at a point (1, 7) touches the circle x...
The given curve is y = x2 + 6 Equation of tangent at (1, 7) is 
(y + 7) = x .1 + 6
⇒ 2x – y + 5 = 0 ...(1)
As given this tangent (1) touches the circle x2 + y2 +16x + 12y + c = 0 at Q
Centre of circle = (– 8, – 6).
(y + 7) = x .1 + 6⇒ 2x – y + 5 = 0 ...(1)
As given this tangent (1) touches the circle x2 + y2 +16x + 12y + c = 0 at Q
Centre of circle = (– 8, – 6).
Then equation of CQ which is perpendicular to (1) and passes through (– 8, – 6) is y + 6 = -(x + 8)
(x + 8)⇒ x + 2y + 20 = 0 ...(2)
Now Q is pt. of intersection of (1) and (2)
∴ Solving eqn (1) & (2) we get x = – 6, y = – 7
∴ Req. pt. is (– 6, – 7).
Now Q is pt. of intersection of (1) and (2)
∴ Solving eqn (1) & (2) we get x = – 6, y = – 7
∴ Req. pt. is (– 6, – 7).
Most Upvoted Answer
Tangent to the curve y = x2 + 6 at a point (1, 7) touches the circle x...
To find the equation of the tangent to the curve y = x^2 - 6 at the point (1,7), we need to find the slope of the tangent at that point. We can do this by taking the derivative of the function:
y' = 2x
At x=1, the slope of the tangent is y'(1) = 2.
So the equation of the tangent is:
y - 7 = 2(x - 1)
Simplifying, we get:
y = 2x + 5
To find the point where this tangent touches the circle x^2 + y^2 - 16x - 12y + c = 0, we substitute y = 2x + 5 into the equation of the circle:
x^2 + (2x+5)^2 - 16x - 12(2x+5) + c = 0
Simplifying, we get:
5x^2 - 72x + c - 107 = 0
We know that the tangent touches the circle at only one point, so the discriminant of this quadratic equation must be zero:
72^2 - 4(5)(c-107) = 0
Solving for c, we get:
c = 2005
So the equation of the circle is:
x^2 + y^2 - 16x - 12y + 2005 = 0
To find the point where the tangent touches the circle, we substitute y = 2x + 5 into the equation of the circle and solve for x:
x^2 + (2x+5)^2 - 16x - 12(2x+5) + 2005 = 0
Simplifying, we get:
5x^2 - 72x + 2005 = 0
Using the quadratic formula, we get:
x = (72 ± √(72^2 - 4(5)(2005))) / (2(5))
x = (72 ± 34√21) / 5
Since the tangent touches the circle at only one point, we choose the solution that is farther away from the point (1,7):
x = (72 + 34√21) / 5
Now we can find y by substituting this value of x into y = 2x + 5:
y = 2((72 + 34√21) / 5) + 5
Simplifying, we get:
y = 144/5 + 68√21/5
So the coordinates of the point where the tangent touches the circle are:
Q = ((72 + 34√21) / 5, 144/5 + 68√21/5)
Therefore, the answer is (2005S) b).
y' = 2x
At x=1, the slope of the tangent is y'(1) = 2.
So the equation of the tangent is:
y - 7 = 2(x - 1)
Simplifying, we get:
y = 2x + 5
To find the point where this tangent touches the circle x^2 + y^2 - 16x - 12y + c = 0, we substitute y = 2x + 5 into the equation of the circle:
x^2 + (2x+5)^2 - 16x - 12(2x+5) + c = 0
Simplifying, we get:
5x^2 - 72x + c - 107 = 0
We know that the tangent touches the circle at only one point, so the discriminant of this quadratic equation must be zero:
72^2 - 4(5)(c-107) = 0
Solving for c, we get:
c = 2005
So the equation of the circle is:
x^2 + y^2 - 16x - 12y + 2005 = 0
To find the point where the tangent touches the circle, we substitute y = 2x + 5 into the equation of the circle and solve for x:
x^2 + (2x+5)^2 - 16x - 12(2x+5) + 2005 = 0
Simplifying, we get:
5x^2 - 72x + 2005 = 0
Using the quadratic formula, we get:
x = (72 ± √(72^2 - 4(5)(2005))) / (2(5))
x = (72 ± 34√21) / 5
Since the tangent touches the circle at only one point, we choose the solution that is farther away from the point (1,7):
x = (72 + 34√21) / 5
Now we can find y by substituting this value of x into y = 2x + 5:
y = 2((72 + 34√21) / 5) + 5
Simplifying, we get:
y = 144/5 + 68√21/5
So the coordinates of the point where the tangent touches the circle are:
Q = ((72 + 34√21) / 5, 144/5 + 68√21/5)
Therefore, the answer is (2005S) b).
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Tangent to the curve y = x2 + 6 at a point (1, 7) touches the circle x2 + y2 + 16x + 12y + c = 0 at a point Q. Then the coordinates of Q are (2005S)a)(–6, –11)b)(–9, –13)c)(–10, –15)d)(–6, –7)Correct answer is option 'D'. Can you explain this answer?
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Tangent to the curve y = x2 + 6 at a point (1, 7) touches the circle x2 + y2 + 16x + 12y + c = 0 at a point Q. Then the coordinates of Q are (2005S)a)(–6, –11)b)(–9, –13)c)(–10, –15)d)(–6, –7)Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Tangent to the curve y = x2 + 6 at a point (1, 7) touches the circle x2 + y2 + 16x + 12y + c = 0 at a point Q. Then the coordinates of Q are (2005S)a)(–6, –11)b)(–9, –13)c)(–10, –15)d)(–6, –7)Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Tangent to the curve y = x2 + 6 at a point (1, 7) touches the circle x2 + y2 + 16x + 12y + c = 0 at a point Q. Then the coordinates of Q are (2005S)a)(–6, –11)b)(–9, –13)c)(–10, –15)d)(–6, –7)Correct answer is option 'D'. Can you explain this answer?.
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