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If O is the origin and OP, OQ are distinct tangents to the circle x2+ y2 + 2gx + 2fy + c = 0, the
circumcentre of the triangle OPQ is-
circumcentre of the triangle OPQ is-
- a)(–g, –f)
- b)(g, f)
- c)(–f, –g)
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
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If O is the origin and OP, OQ are distinct tangents to the circle x2+ ...
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If O is the origin and OP, OQ are distinct tangents to the circle x2+ ...
To find the circumcenter of triangle OPQ, we need to find the intersection point of the perpendicular bisectors of the sides OP and OQ.
First, let's find the equation of the circle given by x^2 + y^2 + 2gx + 2fy + c = 0.
Since O is the origin, we can substitute x = 0 and y = 0 into the equation to get c = 0.
Now, the equation of the circle simplifies to x^2 + y^2 + 2gx + 2fy = 0.
The equation of the line OP can be found by substituting the coordinates of P (x1, y1) into the general equation of a line y - y1 = m(x - x1), where m is the slope of the tangent at P.
Since OP is a tangent, its slope is perpendicular to the radius at point P, which means it is equal to -1/m, where m is the slope of the radius at point P.
The slope of the radius at point P can be found by differentiating the equation of the circle with respect to x and solving for dy/dx at point P.
Differentiating the equation of the circle x^2 + y^2 + 2gx + 2fy = 0 with respect to x, we get:
2x + 2y(dy/dx) + 2g + 2f(dy/dx) = 0
Simplifying, we get:
(dy/dx)(2y + 2f) = -2x - 2g
(dy/dx) = (-2x - 2g)/(2y + 2f)
Since P lies on the circle, we can substitute the coordinates of P into this equation to get the slope of the radius at point P.
Substituting x1 and y1 into the equation, we get:
(dy/dx) = (-2x1 - 2g)/(2y1 + 2f)
The slope of the tangent at point P is the negative reciprocal of this slope, so the slope of the line OP is:
m = -1/((-2x1 - 2g)/(2y1 + 2f))
Simplifying, we get:
m = (2y1 + 2f)/(-2x1 - 2g)
The equation of the line OP is therefore:
y - y1 = (2y1 + 2f)/(-2x1 - 2g)(x - x1)
The equation of the line OQ can be found similarly.
Now, to find the point of intersection of the perpendicular bisectors of OP and OQ, we need to find the midpoint of OP and OQ and the negative reciprocal of their slopes.
The midpoint of OP is ((0 + x1)/2, (0 + y1)/2), which simplifies to (x1/2, y1/2).
The midpoint of OQ is ((0 + x2)/2, (0 + y2)/2), which simplifies to (x2/2, y2/2).
The slope of the perpendicular bisector of OP is the negative reciprocal of the slope of OP, which is (-2x1 - 2g)/(2y1 + 2f).
The slope of the perpendicular bisector of OQ is the negative
First, let's find the equation of the circle given by x^2 + y^2 + 2gx + 2fy + c = 0.
Since O is the origin, we can substitute x = 0 and y = 0 into the equation to get c = 0.
Now, the equation of the circle simplifies to x^2 + y^2 + 2gx + 2fy = 0.
The equation of the line OP can be found by substituting the coordinates of P (x1, y1) into the general equation of a line y - y1 = m(x - x1), where m is the slope of the tangent at P.
Since OP is a tangent, its slope is perpendicular to the radius at point P, which means it is equal to -1/m, where m is the slope of the radius at point P.
The slope of the radius at point P can be found by differentiating the equation of the circle with respect to x and solving for dy/dx at point P.
Differentiating the equation of the circle x^2 + y^2 + 2gx + 2fy = 0 with respect to x, we get:
2x + 2y(dy/dx) + 2g + 2f(dy/dx) = 0
Simplifying, we get:
(dy/dx)(2y + 2f) = -2x - 2g
(dy/dx) = (-2x - 2g)/(2y + 2f)
Since P lies on the circle, we can substitute the coordinates of P into this equation to get the slope of the radius at point P.
Substituting x1 and y1 into the equation, we get:
(dy/dx) = (-2x1 - 2g)/(2y1 + 2f)
The slope of the tangent at point P is the negative reciprocal of this slope, so the slope of the line OP is:
m = -1/((-2x1 - 2g)/(2y1 + 2f))
Simplifying, we get:
m = (2y1 + 2f)/(-2x1 - 2g)
The equation of the line OP is therefore:
y - y1 = (2y1 + 2f)/(-2x1 - 2g)(x - x1)
The equation of the line OQ can be found similarly.
Now, to find the point of intersection of the perpendicular bisectors of OP and OQ, we need to find the midpoint of OP and OQ and the negative reciprocal of their slopes.
The midpoint of OP is ((0 + x1)/2, (0 + y1)/2), which simplifies to (x1/2, y1/2).
The midpoint of OQ is ((0 + x2)/2, (0 + y2)/2), which simplifies to (x2/2, y2/2).
The slope of the perpendicular bisector of OP is the negative reciprocal of the slope of OP, which is (-2x1 - 2g)/(2y1 + 2f).
The slope of the perpendicular bisector of OQ is the negative
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If O is the origin and OP, OQ are distinct tangents to the circle x2+ y2+ 2gx + 2fy + c = 0, thecircumcentre of the triangle OPQ is-a)(–g, –f)b)(g, f)c)(–f, –g)d)None of theseCorrect answer is option 'D'. Can you explain this answer?
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If O is the origin and OP, OQ are distinct tangents to the circle x2+ y2+ 2gx + 2fy + c = 0, thecircumcentre of the triangle OPQ is-a)(–g, –f)b)(g, f)c)(–f, –g)d)None of theseCorrect 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 If O is the origin and OP, OQ are distinct tangents to the circle x2+ y2+ 2gx + 2fy + c = 0, thecircumcentre of the triangle OPQ is-a)(–g, –f)b)(g, f)c)(–f, –g)d)None of theseCorrect 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 If O is the origin and OP, OQ are distinct tangents to the circle x2+ y2+ 2gx + 2fy + c = 0, thecircumcentre of the triangle OPQ is-a)(–g, –f)b)(g, f)c)(–f, –g)d)None of theseCorrect answer is option 'D'. Can you explain this answer?.
If O is the origin and OP, OQ are distinct tangents to the circle x2+ y2+ 2gx + 2fy + c = 0, thecircumcentre of the triangle OPQ is-a)(–g, –f)b)(g, f)c)(–f, –g)d)None of theseCorrect 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 If O is the origin and OP, OQ are distinct tangents to the circle x2+ y2+ 2gx + 2fy + c = 0, thecircumcentre of the triangle OPQ is-a)(–g, –f)b)(g, f)c)(–f, –g)d)None of theseCorrect 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 If O is the origin and OP, OQ are distinct tangents to the circle x2+ y2+ 2gx + 2fy + c = 0, thecircumcentre of the triangle OPQ is-a)(–g, –f)b)(g, f)c)(–f, –g)d)None of theseCorrect answer is option 'D'. Can you explain this answer?.
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