Question

find the equation of the circle two of whose diameters are x+y=6 and x+2y-4=0 and radius is 10 units

Answer 1

Point of intersection of two diameters is the centre.

So, intersection of x+y=6 and x+2y-4=0 is (8,-2)

Circle is (x-8)²+(y+2)²=10²

Answer 2

The equation of the circle with given equations of diameters and radius, is $(x-8)^{2} +(y+2)^{2} =100.$

Given,

For a circle, diameters are:

x + y = 6,

x + 2y - 4 = 0 ⇒ x + 2y = 4, and

radius = 10 units.

To find,

Equation of the circle.

Solution,

First of all, the equation of a circle is given as

$(x-h)^{2} +(y-k)^{2} =r^{2} \hfill ...(1)$

where,

(h, k) = coordinates of the center, and

r = radius of the circle.

Now, we know that the point at which the diameters of a circle intersect is the center of the circle.

Here, the equations of 2 diameters are given, which are

$x + y = 6 \hfill ...(2)$

$x + 2y = 4 \hfill ...(3)$

So, the coordinates of the center can be determined by solving the above equations.

Now, using elimination method, on subtracting equation (1) from (2), we get,

x + 2y - (x + y) = 4 - 6

⇒ x + 2y - x - y = -2

⇒ y = -2.

From (1), we have,

x = 6 - y

⇒ x = 6 - (-2)

⇒ x = 8.

Hence, the coordinates of the center are:

(h, k) = (8, -2), and

r = 10 units.

Substituting the above values in (1), the equation of the circle can be determined as follows.

$(x-8)^{2} +(y-(-2))^{2} =(10)^{2}$

$\implies (x-8)^{2} +(y+2)^{2} =100$

Therefore, the equation of the circle with given equations of diameters and radius, is $(x-8)^{2} +(y+2)^{2} =100.$

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