Question:

Find the centre and radius of the circle x2+y2−8x+10y−12=0. A (4,−5),√53 B (−4,−5),√53 ​C (−4,5),√53 ​D (4,5),√53

Answer:

To solve this problem, we need to use the general equation for a circle:

x2 + y2 + 2gx + 2fy + c = 0

Where (g,f) is the center of the circle, and c is the radius squared.

We can rewrite the given equation as:

x2 + y2 − 8x + 10y − 12 = 0

Comparing this to the general equation, we can see that g = -4, f = -5, and c = 53.

Therefore, the centre of the circle is (-4,-5) and the radius is √53.

The correct answer is B: (-4,-5),√53.

Question:

Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3)

Answer:

The equation of the circle with centre (h,k) and radius r is given by (x-h)2 + (y-k)2 = r2

Given that the centre lies on x-axis, the centre coordinates are (h, 0).

Given that the circle passes through (2,3), substituting x=2 and y=3 in the equation of the circle, we get:

(2-h)2 + (3-0)2 = 52

Solving for h, we get h = -1

Therefore, the equation of the circle is (x+1)2 + y2 = 25

Question:

Find the equation of centre (1,1) and radius √2

Answer:

Answer: Step 1: Identify the equation of a circle. The equation of a circle is (x - h)2 + (y - k)2 = r2, where (h, k) is the center of the circle and r is the radius.

Step 2: Substitute the given values in the equation. Substitute (h, k) = (1, 1) and r = √2 in the equation.

Step 3: Simplify the equation. (x - 1)2 + (y - 1)2 = (√2)2 (x - 1)2 + (y - 1)2 = 2

Hence, the equation of a circle with center (1, 1) and radius √2 is (x - 1)2 + (y - 1)2 = 2.

Question:

Find the centre and radius of the circles. 2x^2+2y^2−x=0

Answer:

1. Rearrange the equation to put it in the form of (x-h)^2 + (y-k)^2 = r^2 2x^2 + 2y^2 - x = 0 2x^2 + 2y^2 - x + 1/4 = 1/4 (2x - 1/2)^2 + 2y^2 = 1/4

2. Identify the values of h, k, and r h = -1/2 k = 0 r = √1/4

3. The centre of the circle is (h, k) = (-1/2, 0) and the radius is r = √1/4

Question:

The equation of the circle passing through the points (4,1),(6,5) and having the centre on the line 4x+y−16=0 is A x^2+y^2−6x−8y+15=0 B 15(x^2+y^2)−94x+18y+55=0 C x^2+y^2−4x−3y=0 D x^2+y^2+6x−4y=0

Answer:

Step 1: Calculate the midpoint of the two given points (4,1) and (6,5). Midpoint = (5,3)

Step 2: Substitute the midpoint in the equation of the line 4x+y−16=0 to calculate the coordinates of the centre of the circle. 4(5)+3−16=0 20+3−16=0 7=0

This equation has no solution, which means that the line 4x+y−16=0 does not pass through the centre of the circle.

Step 3: Use the distance formula to calculate the distance between the two given points (4,1) and (6,5). Distance = √((6−4)^2+(5−1)^2) Distance = √((2)^2+(4)^2) Distance = √(4+16) Distance = √20

Step 4: Use the distance formula to calculate the radius of the circle. Radius = Distance/2 Radius = √20/2 Radius = √10/2 Radius = 5/2

Step 5: Substitute the coordinates of the two points and the radius in the general equation of a circle (x-h)^2+(y-k)^2=r^2 to calculate the equation of the circle. (x−4)^2+(y−1)^2=(5/2)^2 x^2−8x+16+y^2−2y+1=(25/4) x^2−8x+y^2−2y+17/4=0

Hence, the equation of the circle passing through the points (4,1),(6,5) and having the centre on the line 4x+y−16=0 is C x^2+y^2−4x−3y=0.

Question:

Find the equation of the circle passing through the points (2,3) and (−1,1) and whose centre is on the line x−3y−11=0

Answer:

Step 1: Find the centre of the circle.

The centre of the circle lies on the line x−3y−11=0.

We can rearrange the equation to get y = (1/3)x + 11/3.

Therefore, the centre of the circle is (x, (1/3)x + 11/3).

Step 2: Find the radius of the circle.

The radius of the circle is the distance between the centre and one of the given points.

Let’s use the point (2,3).

The distance between the centre and (2,3) is given by the formula d = √((x − 2)2 + ((1/3)x + 11/3 − 3)2).

Substituting x = 2 in the formula, we get d = √((2 − 2)2 + ((1/3)2 + 11/3 − 3)2) = √(0 + (4/3 − 3)2) = √(1/9) = 1/3.

Therefore, the radius of the circle is 1/3.

Step 3: Find the equation of the circle.

The equation of the circle is given by (x − x₀)2 + (y − y₀)2 = r2, where (x₀, y₀) is the centre of the circle and r is the radius of the circle.

Substituting the values of x₀, y₀ and r, we get (x − 2)2 + (y − (1/3)2 + 11/3)2 = (1/3)2.

Simplifying, we get x2 − 4x + 4 + y2 − (2/3)y + (121/9) − (4/3) = 1/9.

Therefore, the equation of the circle is x2 − 4x + 4 + y2 − (2/3)y + (121/9) = 0.

Question:

Equation of circle with center (-a, -b) and radius √a^2−b^2 is. A x^2+y^2−2ax−2by−2b^2=0 B x^2+y^2−2ax−2by+2a^2=0 C x^2+y^2+2ax+2by+2b^2=0 D x^2+y^2−2ax−2by+2b^2=0

Answer:

A) The equation of the circle with center (-a, -b) and radius √a^2−b^2 is D x^2+y^2−2ax−2by+2b^2=0.

Question:

Find the centre and radius of the circle x2+y2−4x−8y−45=0. A (2,6),√63 B (2,4),√65 ​C (2,−4),√66 ​D None

Answer:

Answer: C (2,-4), √66

Question:

Find the equation of circle with centre (2,2) and passes through the point (4,5).

Answer:

Answer: Step 1: Calculate the radius of the circle. Radius = √((x2 - x1)2 + (y2 - y1)2) = √((4 - 2)2 + (5 - 2)2) = √(22 + 32) = √13

Step 2: Write the equation of the circle using the centre and radius. Equation of the circle = (x - 2)2 + (y - 2)2 = 13

Question:

In each of the following Exercises 1 to 5, find the equation of the circle with A (0,2) and radius 2 B centre (−2,3)and radius4 C centre(1​/2,1​/4) and radius1​/12 D centre(1,1)andradius√2

Answer:

A. (x - 0)² + (y - 2)² = 4

B. (x + 2)² + (y - 3)² = 16

C. (x - 1/2)² + (y - 1/4)² = 1/144

D. (x - 1)² + (y - 1)² = 2

Question:

Find the equation of the circle with centre (−2,3) and radius 4

Answer:

Step 1: Identify the center (h,k) of the circle.

Center (h,k) = (−2,3)

Step 2: Identify the radius of the circle.

Radius = 4

Step 3: Write the standard equation of a circle.

(x - h)² + (y - k)² = r²

Step 4: Substitute the values of h, k and r in the equation.

(x + 2)² + (y - 3)² = 4²

Step 5: Simplify the equation.

(x + 2)² + (y - 3)² = 16

Hence, the equation of the circle with centre (−2,3) and radius 4 is (x + 2)² + (y - 3)² = 16.

Question:

Find the equation of the circle with centre (1​/2,1​/4) and radius 1/12

Answer:

Answer: The equation of a circle with centre (h,k) and radius r is (x-h)^2 + (y-k)^2 = r^2

Substituting the given values, we get (x-1/2)^2 + (y-1/4)^2 = (1/12)^2

Simplifying, we get x^2 - x + 1/4 + y^2 - y/2 + 1/16 = 1/144

Rearranging, we get x^2 + y^2 - x - y/2 + 1/16 - 1/144 = 0

Therefore, the equation of the circle is x^2 + y^2 - x - y/2 + 7/144 = 0

Question:

Find the centre and radius of the circle (x+5)^2+(y−3)^2=36

Answer:

1. Centre: (x, y) = (5, 3)
2. Radius: r = 6

Question:

Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinates axes.

Answer:

Answer: Step 1: The equation of the circle can be written as (x-h)2 + (y-k)2 = r2, where (h,k) is the center of the circle and r is the radius of the circle.

Step 2: Since the circle passes through (0,0), the center of the circle must be (0,0).

Step 3: The intercepts of the circle on the x-axis and y-axis are a and b respectively. Thus, the radius of the circle is equal to half of the sum of the intercepts, i.e. r = (a+b)/2

Step 4: Substituting the values of h, k and r in the equation of the circle, the equation of the circle is x2 + y2 = (a+b)2

Question:

Does the point (−2.5,3.5) lie inside outside or on the circle x^2+y^2=25?

Answer:

Step 1: Rewrite the equation of the circle in standard form: (x-h)^2+(y-k)^2=r^2

Answer: (x-0)^2+(y-0)^2=25

Step 2: Substitute the coordinates of the point into the equation: (x-(-2.5))^2+(y-3.5)^2=25

Answer: (x+2.5)^2+(y-3.5)^2=25

Step 3: Simplify the equation: (2.5)^2+(y-3.5)^2=25

Answer: 6.25+(y-3.5)^2=25

Step 4: Solve for y: (y-3.5)^2=18.75

Answer: y-3.5=√18.75

Step 5: Add 3.5 to both sides: y=3.5+√18.75

Answer: y=3.5+√18.75

Step 6: Check to see if the coordinates of the point satisfy the equation: (x+2.5)^2+(y-3.5)^2=25

Answer: (2.5)^2+((3.5+√18.75)-3.5)^2=25

Step 7: Simplify: (2.5)^2+(√18.75)^2=25

Answer: 6.25+18.75=25

Step 8: Solve: 25=25

Answer: The point (−2.5,3.5) lies on the circle x^2+y^2=25.