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Colleges

Triton International College

Location: Subidhanagar, Tinkune

Phone No: 123456790

Courses: BCA, BBA, BIM, BBS, BIT

Triton International College

Location: Subidhanagar, Tinkune

Phone No: 123456790

Courses: BCA, BBA, BIM, BBS, BIT

Circle

18. The equation of the tangent to the circle x2 + y 2 = 5 at (2,1) is [2075]

2x + y = 5

x + y = 5

2x - y = 5

x - y = 5

Solutions
The tangent to a circle equation x2+ y2=a2 at (x1, y1) is xx1+yy1= a2

18. The equation of tangent to circle x2 + y 2 = 9 at (1,1) is [2074]

x + y = 9

x - y = 9

2x + 2y = 9

-x - y = 9

Solutions

18. The equation of tangent to the circle x2 + y2 = 4 at (2,1) is [2072]

x + y = 4

x - y = 4

2x + y = 4

x + 2y = 4

Solutions
The tangent to a circle equation x2+ y2=a2 at (x1, y1) is xx1+yy1= a2

A circle has the equation x² + y² = 16. What is the radius of this circle?

256

8

16

4

Solutions
equation of a circle of center at (h,k) and radius r as (x - h)2 + (y - k)2 = r2
Since circle is in (0,0) (x-0)2 + (y - 0)2 = 16
r²=16 or r = 4

Which of the following equations represents a circle, radius 5 and centre (0,0)?

x² + y² = 25

y² = 5x²

(x − 5)² + (y − 5)² = 0

x² + y² = 5

Solutions
equation of a circle of center at (h,k) and radius r as (x - h)2 + (y - k)2 = r2
Since circle is in (0,0) (x-0)2 + (y - 0)2 = 52
x² + y² = 25

What is the equation of a circle with centre (2,1) and radius 6?

(x − 2)² + (y − 1)² = 6

(x + 2)² + (y + 1)² = 6

(x − 2)² + (y − 1)² = 36

(x + 2)² + (y + 1)² = 36

Solutions
equation of a circle of center at (h,k) and radius r as (x - h)2 + (y - k)2 = r2
Hence (x-2)2 + (y - 4)2 = 36

If a variable circle of radius 4 cuts the circle x² + y² = 1 orthogonally then locus of its centre will be

x² + y² = 16

x² + y² =17

x² + y² – 2x – 4y = 1

2x – 4y + 5 = 0

13. Radius of the circle x 2 + y 2 + 2 y x 3 4 is [2070]

2

1

4

3

Solutions
The equation of the given circle is
x² + y² - 3x - 2y - 3/4 = 0
or, x² - 3x + y² - 2y = 3/4
or, x² - (2 * x * 3/2) + (3/2)² + y² - (2 * y * 1) + 1² = 3/4 + (3/2)² + 1²
or, (x - 3/2)² + (y - 1)² = 3/4 + 9/4 + 1
or, (x - 3/2)² + (y - 1)² = (3 + 9 + 4)/4
or, (x - 3/2)² + (y - 1)² = 16/4
or, (x - 3/2)² + (y - 1)² = 4
or, (x - 3/2)² + (y - 1)² = 2²
Comparing it with the equation of the circle in centre, radius form from the formula part, we get
(a, b) ≡ (3/2, 1) and r = 2 units
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