Question

The lines 2x-3y-5=0 and 3x-4y=7 are diameters of a circle of area 154

Answer 1

Question:

The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle of area 154 sq. units. Find the center and equation of the circle.

Answer:

• Center of the circle is (1,-1).

• The required equation of the circle is,

(x -1)^2 + (y+1)^2 = 7^2

OR

x^2 + y^2 - 2x + 2y = 47

Note:

• Area of the circle of radius"r" is given by; Area = π•r^2
• The equation of the circle with radius "r" and center (h,k) is given by; (x-h)^2 + (y-k)^2 = r^2
• A circle has infinitely many diameters.
• All the diameters of a circle are concurrent at a single point called its centre.
• Lines are said to be concurrent if all of the given lines passes through a single point.
• The approx value of π is 22/7 or 3.14 .

Solution:

It is given that;

The lines 2x - 3y = 5 and 3x - 4y = 7 are the diameters of the circle.

We know that,

The diameters of the circle are concurrent at its centre ,thus the lines

2x - 3y = 5 and 3x - 4y = 7 must intersect at the centre of the circle.

Now,

The equations of the given two diameters are ;

2x - 3y = 5 ---------(1)

3x - 4y = 7 ----------(2)

Now,

Multiplying eq-(1) with 3 , we get;

6x - 9y = 15 ----------(3)

Also,

Multiplying eq-(2) with 2 , we get;

6x - 8y = 14 ----------(4)

Now,

Subtracting eq-(4) from eq-(3), we get;

=> (6x - 9y) - (6x - 8y) = 15 - 14

=> 6x - 9y - 6x + 8y = 1

=> - y = 1

=> y = -1

Now,

Putting y = -1 in eq-(1) , we get;

=> 2x - 3y = 5

=> 2x - 3•(-1) = 5

=> 2x + 3 = 5

=> 2x = 5 - 3

=> 2x = 2

=> x = 2/2

=> x = 1

Hence,

The center of the circle is (1,-1).

Alos,

It is given that , the area of the circle is

154 sq. units.

Thus, we have;

=> π•r^2 = 154

=> (22/7)•r^2 = 154

=> r^2 = 154•(7/22)

=> r^2 = 49

=> r = √49

=> r = 7

Hence,

The radius of the circle is 7 units.

Also,

We know that,

The equation of the circle with radius"r" and center (h,k) is given by;

(x - h)^2 + (y - k)^2 = r^2 .

Here,

The radius of the circle is 7 units and the center of the circle is (1,-1) .

Thus,

The required equation of the circle will be given as;

=> (x -1)^2 + {y-(-1)}^2 = 7^2

=> (x -1)^2 + (y+1)^2 = 7^2

=> x^2 - 2x + 1 + y^2 + 2y + 1 = 49

=> x^2 + y^2 - 2x + 2y + 2 = 49

=> x^2 + y^2 - 2x + 2y = 49 - 2

=> x^2 + y^2 - 2x + 2y = 47

Hence,

The required equation of the circle is,

(x -1)^2 + (y+1)^2 = 7^2

OR

x^2 + y^2 - 2x + 2y = 47

Answer 2

SOLUTION:-

Given:

The lines 2x-3y-5=0 & 3x-4y=7 are Diameters of a circle.

& area= 154.

To find:

The equation of the circle.

Explanation:

We have,

•2x -3y-5=0

•3x -4y=7

Using Substitution Method:

• 2x-3y=5............(1)
• 3x-4y=7............(2)

From equation (1), we get;

=) 2x-3y=5

=) 2x= 5+3y

=) x= 5+3y/2.............(3)

Putting the value of x in equation (2), we get;

$= > 3( \frac{5 + 3y}{2} ) - 4y = 7 \\ \\ = > \frac{15 + 9y}{2} - 4y = 7 \\ \\ = > 15 + 9y - 8y = 14 \\ \\ = > 15 + y = 14 \\ \\ = > y = 14 - 15 \\ \\ = > y = - 1$

Putting the value of y in equation (3), we get;

$= > x = \frac{5 + 3( - 1)}{2} \\ \\ = >x = \frac{5 + ( - 3)}{2} \\ \\ = > x = \frac{5 - 3}{2} \\ \\ = > x = \frac{2}{2} \\ \\ = >x = 1$

Coordinates of the centre of the circle;

=) (1,-1).

•Area of the circle= 154 sq. units.

We know that, area of the circle= πr².

$= > \pi {r}^{2} = 154 \\ \\ = > \frac{22}{7} \times {r}^{2} = 154 \\ \\ = > {r}^{2} = \frac{154 \times 7}{22} \\ \\ = > {r}^{2} = 7 \times 7 \\ \\ = > {r}^{2} = 49 \\ \\ = > r = \sqrt{49} \\ \\ = > r = 7 \: units$

Now,

Equation of the circle:

(x-1)² +(y+1)²= 49

=) x² +1² -2×x×1+ y² +1² +2×y×1=49

=) x² +1-2x +y² +1 +2y =49

=) x² +y² -2x +2y= 49-2

=) x² +y² -2x+2y= 47.

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