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A circle touches a straight line lx + my + n = 0 and cuts the circle x2 + y2 = 9 orthogonally, The locus of centres of such circles is
- a)(lx + my + n)2 = (l2 + m2) (x2 + y2 – 9)
- b)(lx + my – n)2 = (l2 + m2) (x2 + y2– 9)
- c)(lx + my + n)2 = (l2 + m2) (x2+ y2 + 9)
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
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A circle touches a straight line lx + my + n = 0 and cuts the circle x...
Let the equation of the circle is-
x2+y2+2gx+2fy+c = 0
Given, this circle is orthogonal to x2 +y2 −9=0
Condition of orthogonality
x2+y2+2gx+2fy+c = 0
Given, this circle is orthogonal to x2 +y2 −9=0
Condition of orthogonality
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A circle touches a straight line lx + my + n = 0 and cuts the circle x...
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A circle touches a straight line lx + my + n = 0 and cuts the circle x...
The equation of the straight line is lx + my + n = 0. The equation of the circle is x^2 + y^2 - 9 = 0.
To find the locus of the centers of such circles, we need to find the equation that relates the coordinates of the center (x, y) with the coefficients of the straight line (l, m, n).
Let's assume that the center of the circle is (h, k). Since the circle touches the straight line, the distance between the center and the line is equal to the radius of the circle.
The distance between a point (x1, y1) and a line lx + my + n = 0 is given by the formula:
d = |lx1 + my1 + n| / sqrt(l^2 + m^2)
So, the distance between the center (h, k) and the line lx + my + n = 0 is:
r = |lh + mk + n| / sqrt(l^2 + m^2)
Since the circle cuts the circle x^2 + y^2 = 9 orthogonally, the radius of the circle is equal to the radius of the circle x^2 + y^2 = 9, which is 3.
Therefore, we have:
r = 3
|lh + mk + n| / sqrt(l^2 + m^2) = 3
Squaring both sides of the equation, we get:
(lh + mk + n)^2 = 9(l^2 + m^2)
Expanding and simplifying, we get:
l^2h^2 + 2lmhk + m^2k^2 + 2lnh + 2mnk + n^2 = 9l^2 + 9m^2
Rearranging terms, we get:
(l^2 - 9)h^2 + 2lmhk + (m^2 - 9)k^2 + 2lnh + 2mnk + (n^2 - 9l^2 - 9m^2) = 0
Therefore, the locus of the centers of such circles is:
(l^2 - 9)h^2 + 2lmhk + (m^2 - 9)k^2 + 2lnh + 2mnk + (n^2 - 9l^2 - 9m^2) = 0
So, the correct option is:
(lx + my + n)^2 = (l^2 + m^2)(x^2 + y^2)
To find the locus of the centers of such circles, we need to find the equation that relates the coordinates of the center (x, y) with the coefficients of the straight line (l, m, n).
Let's assume that the center of the circle is (h, k). Since the circle touches the straight line, the distance between the center and the line is equal to the radius of the circle.
The distance between a point (x1, y1) and a line lx + my + n = 0 is given by the formula:
d = |lx1 + my1 + n| / sqrt(l^2 + m^2)
So, the distance between the center (h, k) and the line lx + my + n = 0 is:
r = |lh + mk + n| / sqrt(l^2 + m^2)
Since the circle cuts the circle x^2 + y^2 = 9 orthogonally, the radius of the circle is equal to the radius of the circle x^2 + y^2 = 9, which is 3.
Therefore, we have:
r = 3
|lh + mk + n| / sqrt(l^2 + m^2) = 3
Squaring both sides of the equation, we get:
(lh + mk + n)^2 = 9(l^2 + m^2)
Expanding and simplifying, we get:
l^2h^2 + 2lmhk + m^2k^2 + 2lnh + 2mnk + n^2 = 9l^2 + 9m^2
Rearranging terms, we get:
(l^2 - 9)h^2 + 2lmhk + (m^2 - 9)k^2 + 2lnh + 2mnk + (n^2 - 9l^2 - 9m^2) = 0
Therefore, the locus of the centers of such circles is:
(l^2 - 9)h^2 + 2lmhk + (m^2 - 9)k^2 + 2lnh + 2mnk + (n^2 - 9l^2 - 9m^2) = 0
So, the correct option is:
(lx + my + n)^2 = (l^2 + m^2)(x^2 + y^2)
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A circle touches a straight line lx + my + n = 0 and cuts the circle x2+ y2= 9 orthogonally, The locus of centres of such circles isa)(lx + my + n)2= (l2+ m2) (x2+ y2–9)b)(lx + my –n)2= (l2+ m2) (x2+ y2–9)c)(lx + my + n)2= (l2+ m2) (x2+ y2+ 9)d)None of theseCorrect answer is option 'A'. Can you explain this answer?
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A circle touches a straight line lx + my + n = 0 and cuts the circle x2+ y2= 9 orthogonally, The locus of centres of such circles isa)(lx + my + n)2= (l2+ m2) (x2+ y2–9)b)(lx + my –n)2= (l2+ m2) (x2+ y2–9)c)(lx + my + n)2= (l2+ m2) (x2+ y2+ 9)d)None of theseCorrect answer is option 'A'. 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 A circle touches a straight line lx + my + n = 0 and cuts the circle x2+ y2= 9 orthogonally, The locus of centres of such circles isa)(lx + my + n)2= (l2+ m2) (x2+ y2–9)b)(lx + my –n)2= (l2+ m2) (x2+ y2–9)c)(lx + my + n)2= (l2+ m2) (x2+ y2+ 9)d)None of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circle touches a straight line lx + my + n = 0 and cuts the circle x2+ y2= 9 orthogonally, The locus of centres of such circles isa)(lx + my + n)2= (l2+ m2) (x2+ y2–9)b)(lx + my –n)2= (l2+ m2) (x2+ y2–9)c)(lx + my + n)2= (l2+ m2) (x2+ y2+ 9)d)None of theseCorrect answer is option 'A'. Can you explain this answer?.
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