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The centres of those circles which touch the circle, x2 + y2 - 8x - 8y - 4 = 0, externally and also touch the x-axis, lie on:
- a)a circle
- b)an ellipse which is not a circle
- c)a hyperbola
- d)a parabola
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
The centres of those circles which touch the circle, x2 + y2 - 8x - 8y...
x2 + y2 - 8x - 8y - 4 = 0
Centre (4, 4)
Radius =
= 6
Let centre of the circle is (h, k)
= (6 + k)
(h - 4)2 + (k - 4)2 = (6 + k)2
h2 - 8h + 16 + k2 - 8k + 16 = 36 + k2 + 12k
h2 - 8h - 20k - 4 = 0
x2 - 8x - 20y - 4 = 0
Which is an equation of parabola
Centre (4, 4)
Radius =
= 6Let centre of the circle is (h, k)
= (6 + k)(h - 4)2 + (k - 4)2 = (6 + k)2
h2 - 8h + 16 + k2 - 8k + 16 = 36 + k2 + 12k
h2 - 8h - 20k - 4 = 0
x2 - 8x - 20y - 4 = 0
Which is an equation of parabola
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The centres of those circles which touch the circle, x2 + y2 - 8x - 8y...
Exam Category: JEE
Problem:
The centers of those circles which touch the circle, x² + y² - 8x - 8y - 4 = 0, externally and also touch the x-axis, lie on:
a) a circle
b) an ellipse which is not a circle
c) a hyperbola
d) a parabola
Solution:
To find the centers of the circles that satisfy the given conditions, we can start by analyzing the given equation of the circle: x² + y² - 8x - 8y - 4 = 0.
Step 1: Completing the Square
To simplify the equation, we can complete the square by rearranging the terms:
(x² - 8x) + (y² - 8y) = 4
Take half of the coefficient of x (-8) and square it:
(x² - 8x + 16) + (y² - 8y) = 4 + 16
Similarly, take half of the coefficient of y (-8) and square it:
(x² - 8x + 16) + (y² - 8y + 16) = 4 + 16 + 16
Simplifying further:
(x - 4)² + (y - 4)² = 36
Step 2: Plotting the Circle
The equation (x - 4)² + (y - 4)² = 36 represents a circle with center (4, 4) and radius 6.
We can plot this circle on a coordinate plane.
Step 3: Analysis of the Circles
Now, let's consider the circles that touch the given circle externally and also touch the x-axis.
Key Point 1:
The centers of these circles must lie on a line that is perpendicular to the x-axis and passes through the center of the given circle (4, 4).
Key Point 2:
The distance between the centers of the given circle and the externally touching circles is equal to the sum of their radii.
Step 4: Finding the Centers
Let's consider a point (h, 0) on the x-axis, which represents a possible center of one of the externally touching circles.
Using Key Point 2, we can write the equation:
√((h - 4)² + (0 - 4)²) = 6 + r
Simplifying this equation:
√((h - 4)² + 16) = 6 + r
Squaring both sides:
(h - 4)² + 16 = (6 + r)²
h² - 8h + 16 + 16 = r² + 12r + 36
h² - 8h - r² - 12r - 36 = 0
Step 5: Analyzing the Equation
The equation h² - 8h - r² - 12r - 36 = 0 represents a quadratic equation in two variables, h and r.
Key Point 3:
The equation h² - 8h - r² - 12r -
Problem:
The centers of those circles which touch the circle, x² + y² - 8x - 8y - 4 = 0, externally and also touch the x-axis, lie on:
a) a circle
b) an ellipse which is not a circle
c) a hyperbola
d) a parabola
Solution:
To find the centers of the circles that satisfy the given conditions, we can start by analyzing the given equation of the circle: x² + y² - 8x - 8y - 4 = 0.
Step 1: Completing the Square
To simplify the equation, we can complete the square by rearranging the terms:
(x² - 8x) + (y² - 8y) = 4
Take half of the coefficient of x (-8) and square it:
(x² - 8x + 16) + (y² - 8y) = 4 + 16
Similarly, take half of the coefficient of y (-8) and square it:
(x² - 8x + 16) + (y² - 8y + 16) = 4 + 16 + 16
Simplifying further:
(x - 4)² + (y - 4)² = 36
Step 2: Plotting the Circle
The equation (x - 4)² + (y - 4)² = 36 represents a circle with center (4, 4) and radius 6.
We can plot this circle on a coordinate plane.
Step 3: Analysis of the Circles
Now, let's consider the circles that touch the given circle externally and also touch the x-axis.
Key Point 1:
The centers of these circles must lie on a line that is perpendicular to the x-axis and passes through the center of the given circle (4, 4).
Key Point 2:
The distance between the centers of the given circle and the externally touching circles is equal to the sum of their radii.
Step 4: Finding the Centers
Let's consider a point (h, 0) on the x-axis, which represents a possible center of one of the externally touching circles.
Using Key Point 2, we can write the equation:
√((h - 4)² + (0 - 4)²) = 6 + r
Simplifying this equation:
√((h - 4)² + 16) = 6 + r
Squaring both sides:
(h - 4)² + 16 = (6 + r)²
h² - 8h + 16 + 16 = r² + 12r + 36
h² - 8h - r² - 12r - 36 = 0
Step 5: Analyzing the Equation
The equation h² - 8h - r² - 12r - 36 = 0 represents a quadratic equation in two variables, h and r.
Key Point 3:
The equation h² - 8h - r² - 12r -
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The centres of those circles which touch the circle, x2 + y2 - 8x - 8y - 4 = 0, externally and also touch the x-axis, lie on:a)a circleb)an ellipse which is not a circlec)a hyperbolad)a parabolaCorrect answer is option 'D'. Can you explain this answer?
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The centres of those circles which touch the circle, x2 + y2 - 8x - 8y - 4 = 0, externally and also touch the x-axis, lie on:a)a circleb)an ellipse which is not a circlec)a hyperbolad)a parabolaCorrect 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 The centres of those circles which touch the circle, x2 + y2 - 8x - 8y - 4 = 0, externally and also touch the x-axis, lie on:a)a circleb)an ellipse which is not a circlec)a hyperbolad)a parabolaCorrect 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 The centres of those circles which touch the circle, x2 + y2 - 8x - 8y - 4 = 0, externally and also touch the x-axis, lie on:a)a circleb)an ellipse which is not a circlec)a hyperbolad)a parabolaCorrect answer is option 'D'. Can you explain this answer?.
The centres of those circles which touch the circle, x2 + y2 - 8x - 8y - 4 = 0, externally and also touch the x-axis, lie on:a)a circleb)an ellipse which is not a circlec)a hyperbolad)a parabolaCorrect 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 The centres of those circles which touch the circle, x2 + y2 - 8x - 8y - 4 = 0, externally and also touch the x-axis, lie on:a)a circleb)an ellipse which is not a circlec)a hyperbolad)a parabolaCorrect 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 The centres of those circles which touch the circle, x2 + y2 - 8x - 8y - 4 = 0, externally and also touch the x-axis, lie on:a)a circleb)an ellipse which is not a circlec)a hyperbolad)a parabolaCorrect answer is option 'D'. Can you explain this answer?.
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The centres of those circles which touch the circle, x2 + y2 - 8x - 8y - 4 = 0, externally and also touch the x-axis, lie on:a)a circleb)an ellipse which is not a circlec)a hyperbolad)a parabolaCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for The centres of those circles which touch the circle, x2 + y2 - 8x - 8y - 4 = 0, externally and also touch the x-axis, lie on:a)a circleb)an ellipse which is not a circlec)a hyperbolad)a parabolaCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of The centres of those circles which touch the circle, x2 + y2 - 8x - 8y - 4 = 0, externally and also touch the x-axis, lie on:a)a circleb)an ellipse which is not a circlec)a hyperbolad)a parabolaCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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