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Let E1E2 and F1F2 be the chords of S passing through the point P0(1, 1) and parallel to the x-axis and the yaxis, respectively. Let G1G2 be the chord of S passing through P0 and having slope –1. Let the tangents to S at E1 and E2 meet at E3, the tangents to S at F1 and F2 meet at F3, and the tangents to S at G1 and G2 meet at G3. Then, the points E3, F3, and G3 lie on the curve
- a)x + y = 4
- b)(x – 4)2 + (y – 4)2 = 16
- c)(x – 4) (y – 4) = 4
- d)xy = 4
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Let E1E2 and F1F2 be the chords of S passing through the point P0(1, 1...
Clearly they lie on x + y = 4
Most Upvoted Answer
Let E1E2 and F1F2 be the chords of S passing through the point P0(1, 1...
Let the equation of the circle S be (x-a)^2 + (y-b)^2 = r^2, where (a,b) is the center of the circle and r is the radius.
Since the chord E1E2 is parallel to the x-axis and passes through P0(1, 1), the y-coordinate of both E1 and E2 is 1. Therefore, the equation of the chord E1E2 is y = 1.
Since the chord F1F2 is parallel to the y-axis and passes through P0(1, 1), the x-coordinate of both F1 and F2 is 1. Therefore, the equation of the chord F1F2 is x = 1.
Now, let's find the equation of the chord G1G2 passing through P0(1, 1) and having slope m.
The equation of a line passing through point (x1, y1) with slope m is given by y - y1 = m(x - x1).
Substituting P0(1, 1) into the equation, we have y - 1 = m(x - 1).
Since this line passes through P0, the coordinates (x, y) of any point on this line satisfy the equation of the circle S.
Substituting (x, y) into the equation of S, we have (x-a)^2 + (y-b)^2 = r^2.
Substituting y = mx + (1 - m) into the equation, we have (x-a)^2 + (mx + (1 - m) - b)^2 = r^2.
Expanding and rearranging the equation, we have (1 + m^2)x^2 + (2(1 - m)b - 2a)x + (a^2 + (1 - m)^2 - r^2 + b^2) = 0.
This is a quadratic equation in x. Since G1G2 is a chord of S, this equation has two distinct real solutions for x.
Since P0(1, 1) lies on the line G1G2, it means that one of the solutions for x is x = 1.
Substituting x = 1 into the equation, we have (1 + m^2) + (2(1 - m)b - 2a) + (a^2 + (1 - m)^2 - r^2 + b^2) = 0.
Simplifying the equation, we have a^2 + m^2 - 2am + b^2 - 2bm + 2b - r^2 + 2m - 1 = 0.
Since this equation holds for any value of m, each coefficient of m must be zero.
Therefore, we have the following system of equations:
-2a - 2b = 0, which gives a + b = 0.
2 - 2a - 2b = 0, which gives a + b = 1.
a^2 + b^2 - r^2 - 1 = 0.
Solving the first two equations, we have a = -b and a + b = 1.
Substituting a = -b into the second equation, we have -b + b = 1, which gives b =
Since the chord E1E2 is parallel to the x-axis and passes through P0(1, 1), the y-coordinate of both E1 and E2 is 1. Therefore, the equation of the chord E1E2 is y = 1.
Since the chord F1F2 is parallel to the y-axis and passes through P0(1, 1), the x-coordinate of both F1 and F2 is 1. Therefore, the equation of the chord F1F2 is x = 1.
Now, let's find the equation of the chord G1G2 passing through P0(1, 1) and having slope m.
The equation of a line passing through point (x1, y1) with slope m is given by y - y1 = m(x - x1).
Substituting P0(1, 1) into the equation, we have y - 1 = m(x - 1).
Since this line passes through P0, the coordinates (x, y) of any point on this line satisfy the equation of the circle S.
Substituting (x, y) into the equation of S, we have (x-a)^2 + (y-b)^2 = r^2.
Substituting y = mx + (1 - m) into the equation, we have (x-a)^2 + (mx + (1 - m) - b)^2 = r^2.
Expanding and rearranging the equation, we have (1 + m^2)x^2 + (2(1 - m)b - 2a)x + (a^2 + (1 - m)^2 - r^2 + b^2) = 0.
This is a quadratic equation in x. Since G1G2 is a chord of S, this equation has two distinct real solutions for x.
Since P0(1, 1) lies on the line G1G2, it means that one of the solutions for x is x = 1.
Substituting x = 1 into the equation, we have (1 + m^2) + (2(1 - m)b - 2a) + (a^2 + (1 - m)^2 - r^2 + b^2) = 0.
Simplifying the equation, we have a^2 + m^2 - 2am + b^2 - 2bm + 2b - r^2 + 2m - 1 = 0.
Since this equation holds for any value of m, each coefficient of m must be zero.
Therefore, we have the following system of equations:
-2a - 2b = 0, which gives a + b = 0.
2 - 2a - 2b = 0, which gives a + b = 1.
a^2 + b^2 - r^2 - 1 = 0.
Solving the first two equations, we have a = -b and a + b = 1.
Substituting a = -b into the second equation, we have -b + b = 1, which gives b =
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Let E1E2 and F1F2 be the chords of S passing through the point P0(1, 1) and parallel to the x-axis and the yaxis, respectively. Let G1G2 be the chord of S passing through P0 and having slope –1. Let the tangents to S at E1 and E2 meet at E3, the tangents to S at F1 and F2 meet at F3, and the tangents to S at G1 and G2 meet at G3. Then, the points E3, F3, and G3 lie on the curvea)x + y = 4b)(x – 4)2 + (y – 4)2 = 16c)(x – 4) (y – 4) = 4d)xy = 4Correct answer is option 'A'. Can you explain this answer?
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Let E1E2 and F1F2 be the chords of S passing through the point P0(1, 1) and parallel to the x-axis and the yaxis, respectively. Let G1G2 be the chord of S passing through P0 and having slope –1. Let the tangents to S at E1 and E2 meet at E3, the tangents to S at F1 and F2 meet at F3, and the tangents to S at G1 and G2 meet at G3. Then, the points E3, F3, and G3 lie on the curvea)x + y = 4b)(x – 4)2 + (y – 4)2 = 16c)(x – 4) (y – 4) = 4d)xy = 4Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let E1E2 and F1F2 be the chords of S passing through the point P0(1, 1) and parallel to the x-axis and the yaxis, respectively. Let G1G2 be the chord of S passing through P0 and having slope –1. Let the tangents to S at E1 and E2 meet at E3, the tangents to S at F1 and F2 meet at F3, and the tangents to S at G1 and G2 meet at G3. Then, the points E3, F3, and G3 lie on the curvea)x + y = 4b)(x – 4)2 + (y – 4)2 = 16c)(x – 4) (y – 4) = 4d)xy = 4Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let E1E2 and F1F2 be the chords of S passing through the point P0(1, 1) and parallel to the x-axis and the yaxis, respectively. Let G1G2 be the chord of S passing through P0 and having slope –1. Let the tangents to S at E1 and E2 meet at E3, the tangents to S at F1 and F2 meet at F3, and the tangents to S at G1 and G2 meet at G3. Then, the points E3, F3, and G3 lie on the curvea)x + y = 4b)(x – 4)2 + (y – 4)2 = 16c)(x – 4) (y – 4) = 4d)xy = 4Correct answer is option 'A'. Can you explain this answer?.
Let E1E2 and F1F2 be the chords of S passing through the point P0(1, 1) and parallel to the x-axis and the yaxis, respectively. Let G1G2 be the chord of S passing through P0 and having slope –1. Let the tangents to S at E1 and E2 meet at E3, the tangents to S at F1 and F2 meet at F3, and the tangents to S at G1 and G2 meet at G3. Then, the points E3, F3, and G3 lie on the curvea)x + y = 4b)(x – 4)2 + (y – 4)2 = 16c)(x – 4) (y – 4) = 4d)xy = 4Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let E1E2 and F1F2 be the chords of S passing through the point P0(1, 1) and parallel to the x-axis and the yaxis, respectively. Let G1G2 be the chord of S passing through P0 and having slope –1. Let the tangents to S at E1 and E2 meet at E3, the tangents to S at F1 and F2 meet at F3, and the tangents to S at G1 and G2 meet at G3. Then, the points E3, F3, and G3 lie on the curvea)x + y = 4b)(x – 4)2 + (y – 4)2 = 16c)(x – 4) (y – 4) = 4d)xy = 4Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let E1E2 and F1F2 be the chords of S passing through the point P0(1, 1) and parallel to the x-axis and the yaxis, respectively. Let G1G2 be the chord of S passing through P0 and having slope –1. Let the tangents to S at E1 and E2 meet at E3, the tangents to S at F1 and F2 meet at F3, and the tangents to S at G1 and G2 meet at G3. Then, the points E3, F3, and G3 lie on the curvea)x + y = 4b)(x – 4)2 + (y – 4)2 = 16c)(x – 4) (y – 4) = 4d)xy = 4Correct answer is option 'A'. Can you explain this answer?.
Solutions for Let E1E2 and F1F2 be the chords of S passing through the point P0(1, 1) and parallel to the x-axis and the yaxis, respectively. Let G1G2 be the chord of S passing through P0 and having slope –1. Let the tangents to S at E1 and E2 meet at E3, the tangents to S at F1 and F2 meet at F3, and the tangents to S at G1 and G2 meet at G3. Then, the points E3, F3, and G3 lie on the curvea)x + y = 4b)(x – 4)2 + (y – 4)2 = 16c)(x – 4) (y – 4) = 4d)xy = 4Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Here you can find the meaning of Let E1E2 and F1F2 be the chords of S passing through the point P0(1, 1) and parallel to the x-axis and the yaxis, respectively. Let G1G2 be the chord of S passing through P0 and having slope –1. Let the tangents to S at E1 and E2 meet at E3, the tangents to S at F1 and F2 meet at F3, and the tangents to S at G1 and G2 meet at G3. Then, the points E3, F3, and G3 lie on the curvea)x + y = 4b)(x – 4)2 + (y – 4)2 = 16c)(x – 4) (y – 4) = 4d)xy = 4Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Let E1E2 and F1F2 be the chords of S passing through the point P0(1, 1) and parallel to the x-axis and the yaxis, respectively. Let G1G2 be the chord of S passing through P0 and having slope –1. Let the tangents to S at E1 and E2 meet at E3, the tangents to S at F1 and F2 meet at F3, and the tangents to S at G1 and G2 meet at G3. Then, the points E3, F3, and G3 lie on the curvea)x + y = 4b)(x – 4)2 + (y – 4)2 = 16c)(x – 4) (y – 4) = 4d)xy = 4Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Let E1E2 and F1F2 be the chords of S passing through the point P0(1, 1) and parallel to the x-axis and the yaxis, respectively. Let G1G2 be the chord of S passing through P0 and having slope –1. Let the tangents to S at E1 and E2 meet at E3, the tangents to S at F1 and F2 meet at F3, and the tangents to S at G1 and G2 meet at G3. Then, the points E3, F3, and G3 lie on the curvea)x + y = 4b)(x – 4)2 + (y – 4)2 = 16c)(x – 4) (y – 4) = 4d)xy = 4Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Let E1E2 and F1F2 be the chords of S passing through the point P0(1, 1) and parallel to the x-axis and the yaxis, respectively. Let G1G2 be the chord of S passing through P0 and having slope –1. Let the tangents to S at E1 and E2 meet at E3, the tangents to S at F1 and F2 meet at F3, and the tangents to S at G1 and G2 meet at G3. Then, the points E3, F3, and G3 lie on the curvea)x + y = 4b)(x – 4)2 + (y – 4)2 = 16c)(x – 4) (y – 4) = 4d)xy = 4Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Let E1E2 and F1F2 be the chords of S passing through the point P0(1, 1) and parallel to the x-axis and the yaxis, respectively. Let G1G2 be the chord of S passing through P0 and having slope –1. Let the tangents to S at E1 and E2 meet at E3, the tangents to S at F1 and F2 meet at F3, and the tangents to S at G1 and G2 meet at G3. Then, the points E3, F3, and G3 lie on the curvea)x + y = 4b)(x – 4)2 + (y – 4)2 = 16c)(x – 4) (y – 4) = 4d)xy = 4Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
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