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Equation of the tangent to the circle, at the point (1, −1), whose centre is the point of intersection of the straight lines x−y=1 and 2x+y=3 is :
- a)4x+y−3=0
- b)x+4y+3=0
- c)3x−y−4=0
- d)x−3y−4=0
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
Equation of the tangent to the circle, at the point (1, −1), who...
Since the point (1, has been cut off in your question, I cannot provide the exact equation of the tangent to the circle at that point. However, I can provide you with the general steps to find the equation of the tangent.
To find the equation of the tangent to a circle at a specific point, you need to know the coordinates of the center of the circle and the radius. Let's assume the center of the circle is (h, k) and the radius is r.
1. Determine the slope of the radius from the center of the circle to the given point using the formula: slope = (y - k) / (x - h), where (x, y) is the given point.
2. The slope of the tangent line is the negative reciprocal of the slope of the radius. So, find the negative reciprocal of the slope calculated in step 1.
3. Use the point-slope form of the equation of a line, y - y1 = m(x - x1), where m is the slope from step 2 and (x1, y1) is the given point.
4. Simplify the equation obtained in step 3 to get the final equation of the tangent to the circle.
Please provide the missing coordinate so that I can give you the specific equation of the tangent.
To find the equation of the tangent to a circle at a specific point, you need to know the coordinates of the center of the circle and the radius. Let's assume the center of the circle is (h, k) and the radius is r.
1. Determine the slope of the radius from the center of the circle to the given point using the formula: slope = (y - k) / (x - h), where (x, y) is the given point.
2. The slope of the tangent line is the negative reciprocal of the slope of the radius. So, find the negative reciprocal of the slope calculated in step 1.
3. Use the point-slope form of the equation of a line, y - y1 = m(x - x1), where m is the slope from step 2 and (x1, y1) is the given point.
4. Simplify the equation obtained in step 3 to get the final equation of the tangent to the circle.
Please provide the missing coordinate so that I can give you the specific equation of the tangent.
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Equation of the tangent to the circle, at the point (1, −1), whose centre is the point of intersection of the straight lines x−y=1 and 2x+y=3 is :a)4x+y−3=0b)x+4y+3=0c)3x−y−4=0d)x−3y−4=0Correct answer is option 'B'. Can you explain this answer?
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Equation of the tangent to the circle, at the point (1, −1), whose centre is the point of intersection of the straight lines x−y=1 and 2x+y=3 is :a)4x+y−3=0b)x+4y+3=0c)3x−y−4=0d)x−3y−4=0Correct answer is option 'B'. 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 Equation of the tangent to the circle, at the point (1, −1), whose centre is the point of intersection of the straight lines x−y=1 and 2x+y=3 is :a)4x+y−3=0b)x+4y+3=0c)3x−y−4=0d)x−3y−4=0Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Equation of the tangent to the circle, at the point (1, −1), whose centre is the point of intersection of the straight lines x−y=1 and 2x+y=3 is :a)4x+y−3=0b)x+4y+3=0c)3x−y−4=0d)x−3y−4=0Correct answer is option 'B'. Can you explain this answer?.
Equation of the tangent to the circle, at the point (1, −1), whose centre is the point of intersection of the straight lines x−y=1 and 2x+y=3 is :a)4x+y−3=0b)x+4y+3=0c)3x−y−4=0d)x−3y−4=0Correct answer is option 'B'. 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 Equation of the tangent to the circle, at the point (1, −1), whose centre is the point of intersection of the straight lines x−y=1 and 2x+y=3 is :a)4x+y−3=0b)x+4y+3=0c)3x−y−4=0d)x−3y−4=0Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Equation of the tangent to the circle, at the point (1, −1), whose centre is the point of intersection of the straight lines x−y=1 and 2x+y=3 is :a)4x+y−3=0b)x+4y+3=0c)3x−y−4=0d)x−3y−4=0Correct answer is option 'B'. Can you explain this answer?.
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