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Let the circles S1 ≡ x2 + y2 – 4x – 8y + 4 = 0 and S2 be its image in the line y = x, the equation of the circle touching y = x at (1, 1) and orthogonal to S2 is
- a)x2 + y2 + x – 5y + 2 = 0
- b)x2 + y2 = 2
- c)x2 + y2 + x – 5y − 2 = 0
- d)(x – 3)2 + (y – 2)2 = 5
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Let the circles S1 x2 + y2 4x 8y + 4 = 0 and S2 be its image in the...
Equation of circle touching y = x at (1, 1) can be taken as
As this is orthogonal to S2
∴ required equation of circle is x2 + y2 + x – 5y + 2 = 0.
Most Upvoted Answer
Let the circles S1 x2 + y2 4x 8y + 4 = 0 and S2 be its image in the...
To find the equation of the circle touching the line y = x at (1, 1) and orthogonal to S2, we can follow the following steps:
Step 1: Find the equation of S1
The given equation of the circle S1 is x^2 + y^2 + 4x + 8y + 4 = 0. We can rewrite this equation in the standard form as (x + 2)^2 + (y + 4)^2 = 0.
Step 2: Find the image of S1 in the line y = x
To find the image of S1 in the line y = x, we substitute y = x in the equation of S1. This gives us (x + 2)^2 + (x + 4)^2 = 0.
Step 3: Find the equation of the circle touching y = x at (1, 1) and orthogonal to S2
The circle that touches y = x at (1, 1) will have its center on the line y = x and will be orthogonal to S2.
We know that the tangent to a circle is perpendicular to the radius at the point of contact. Since the circle is orthogonal to S2, the radius of the circle will be perpendicular to the radius of S2 at the point of contact.
The equation of the line y = x can be written as x - y = 0.
The slope of the line y = x is 1, so the slope of the radius of the circle will be -1 (negative reciprocal).
Since the radius passes through (1, 1), we can write the equation of the radius as y - 1 = -1(x - 1).
The equation of the circle touching y = x at (1, 1) and orthogonal to S2 can be written as (x - a)^2 + (y - b)^2 = r^2, where (a, b) is the center of the circle and r is the radius.
Step 4: Substitute the values of (a, b) and r in the equation
Substituting the values of (a, b) = (1, 1) and r = distance between (1, 1) and the line x - y = 0 in the equation, we get:
(x - 1)^2 + (y - 1)^2 = (√2)^2
(x - 1)^2 + (y - 1)^2 = 2
Therefore, the equation of the circle touching y = x at (1, 1) and orthogonal to S2 is x^2 + y^2 - 2x - 2y = 0, which is option A.
Step 1: Find the equation of S1
The given equation of the circle S1 is x^2 + y^2 + 4x + 8y + 4 = 0. We can rewrite this equation in the standard form as (x + 2)^2 + (y + 4)^2 = 0.
Step 2: Find the image of S1 in the line y = x
To find the image of S1 in the line y = x, we substitute y = x in the equation of S1. This gives us (x + 2)^2 + (x + 4)^2 = 0.
Step 3: Find the equation of the circle touching y = x at (1, 1) and orthogonal to S2
The circle that touches y = x at (1, 1) will have its center on the line y = x and will be orthogonal to S2.
We know that the tangent to a circle is perpendicular to the radius at the point of contact. Since the circle is orthogonal to S2, the radius of the circle will be perpendicular to the radius of S2 at the point of contact.
The equation of the line y = x can be written as x - y = 0.
The slope of the line y = x is 1, so the slope of the radius of the circle will be -1 (negative reciprocal).
Since the radius passes through (1, 1), we can write the equation of the radius as y - 1 = -1(x - 1).
The equation of the circle touching y = x at (1, 1) and orthogonal to S2 can be written as (x - a)^2 + (y - b)^2 = r^2, where (a, b) is the center of the circle and r is the radius.
Step 4: Substitute the values of (a, b) and r in the equation
Substituting the values of (a, b) = (1, 1) and r = distance between (1, 1) and the line x - y = 0 in the equation, we get:
(x - 1)^2 + (y - 1)^2 = (√2)^2
(x - 1)^2 + (y - 1)^2 = 2
Therefore, the equation of the circle touching y = x at (1, 1) and orthogonal to S2 is x^2 + y^2 - 2x - 2y = 0, which is option A.
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Let the circles S1 x2 + y2 4x 8y + 4 = 0 and S2 be its image in the line y = x, the equation of the circle touching y = x at (1, 1) and orthogonal to S2 isa)x2 + y2 + x 5y + 2 = 0 b)x2 + y2 = 2 c)x2 + y2 + x 5y 2 = 0 d)(x 3)2 + (y 2)2 = 5 Correct answer is option 'A'. Can you explain this answer?
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Let the circles S1 x2 + y2 4x 8y + 4 = 0 and S2 be its image in the line y = x, the equation of the circle touching y = x at (1, 1) and orthogonal to S2 isa)x2 + y2 + x 5y + 2 = 0 b)x2 + y2 = 2 c)x2 + y2 + x 5y 2 = 0 d)(x 3)2 + (y 2)2 = 5 Correct 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 the circles S1 x2 + y2 4x 8y + 4 = 0 and S2 be its image in the line y = x, the equation of the circle touching y = x at (1, 1) and orthogonal to S2 isa)x2 + y2 + x 5y + 2 = 0 b)x2 + y2 = 2 c)x2 + y2 + x 5y 2 = 0 d)(x 3)2 + (y 2)2 = 5 Correct 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 the circles S1 x2 + y2 4x 8y + 4 = 0 and S2 be its image in the line y = x, the equation of the circle touching y = x at (1, 1) and orthogonal to S2 isa)x2 + y2 + x 5y + 2 = 0 b)x2 + y2 = 2 c)x2 + y2 + x 5y 2 = 0 d)(x 3)2 + (y 2)2 = 5 Correct answer is option 'A'. Can you explain this answer?.
Let the circles S1 x2 + y2 4x 8y + 4 = 0 and S2 be its image in the line y = x, the equation of the circle touching y = x at (1, 1) and orthogonal to S2 isa)x2 + y2 + x 5y + 2 = 0 b)x2 + y2 = 2 c)x2 + y2 + x 5y 2 = 0 d)(x 3)2 + (y 2)2 = 5 Correct 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 the circles S1 x2 + y2 4x 8y + 4 = 0 and S2 be its image in the line y = x, the equation of the circle touching y = x at (1, 1) and orthogonal to S2 isa)x2 + y2 + x 5y + 2 = 0 b)x2 + y2 = 2 c)x2 + y2 + x 5y 2 = 0 d)(x 3)2 + (y 2)2 = 5 Correct 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 the circles S1 x2 + y2 4x 8y + 4 = 0 and S2 be its image in the line y = x, the equation of the circle touching y = x at (1, 1) and orthogonal to S2 isa)x2 + y2 + x 5y + 2 = 0 b)x2 + y2 = 2 c)x2 + y2 + x 5y 2 = 0 d)(x 3)2 + (y 2)2 = 5 Correct answer is option 'A'. Can you explain this answer?.
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Let the circles S1 x2 + y2 4x 8y + 4 = 0 and S2 be its image in the line y = x, the equation of the circle touching y = x at (1, 1) and orthogonal to S2 isa)x2 + y2 + x 5y + 2 = 0 b)x2 + y2 = 2 c)x2 + y2 + x 5y 2 = 0 d)(x 3)2 + (y 2)2 = 5 Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Let the circles S1 x2 + y2 4x 8y + 4 = 0 and S2 be its image in the line y = x, the equation of the circle touching y = x at (1, 1) and orthogonal to S2 isa)x2 + y2 + x 5y + 2 = 0 b)x2 + y2 = 2 c)x2 + y2 + x 5y 2 = 0 d)(x 3)2 + (y 2)2 = 5 Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Let the circles S1 x2 + y2 4x 8y + 4 = 0 and S2 be its image in the line y = x, the equation of the circle touching y = x at (1, 1) and orthogonal to S2 isa)x2 + y2 + x 5y + 2 = 0 b)x2 + y2 = 2 c)x2 + y2 + x 5y 2 = 0 d)(x 3)2 + (y 2)2 = 5 Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Let the circles S1 x2 + y2 4x 8y + 4 = 0 and S2 be its image in the line y = x, the equation of the circle touching y = x at (1, 1) and orthogonal to S2 isa)x2 + y2 + x 5y + 2 = 0 b)x2 + y2 = 2 c)x2 + y2 + x 5y 2 = 0 d)(x 3)2 + (y 2)2 = 5 Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
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