Question

Consider the equation of a circle x²-2x+y²-4y-4=0. If the line 2x-y+a=0 is its diameter. Then find the value of a.​

i need full solution.

Answer 1

Consider the equation of the circle.

$\small\longrightarrow x^2-2x+y^2-4y-4=0$

$\small\longrightarrow x^2-2x+y^2-4y=4$

$\small\longrightarrow x^2-2x+1+y^2-4y+4=4+5$

$\small\longrightarrow(x^2-2x+1)+(y^2-4y+4)=9$

$\small\longrightarrow(x-1)^2+(y-2)^2=3^2$

From this equation we can understand that the center of the circle is (1, 2) and the radius is 3 units.

[Recall that the general equation of a circle of radius r with center at (h, k) is (x - h)² + (y - k)² = 1.]

Since the line 2x - y + a = 0 is a diameter of the circle, (1, 2) is a point on the line as diameter passes through the center.

So taking (x, y) = (1, 2),

$\small\longrightarrow2(1)-(2)+a=0$

$\small\text{ \longrightarrow\underline{\underline{a=0}} }$

Hence the value of a is 0.

Answer 2

A circle x²-2x+y²-4y-4=0 and  line 2x-y+a=0 is its diameter then value of a is 0

Step 1:

Rewrite Equation of the circle in (x - h)² + (y - k)² = r² form

where (h, k) is center and r is radius

x² - 2x + y² - 4y - 4 = 0

=> (x - 1)² - 1 + (y - 2)² - 4 - 4 = 0

=> (x - 1)² + (y - 2)² = 9

=> (x - 1)² + (y - 2)² =  3²

Step 2:

From the equation of the circle , Find the center

h = 1 , k = 2

=> Center = ( 1 , 2)

Step 3:

Diameter passes through center Hence (1 , 2)  will satisfy equation of the diameter 2x - y + a = 0  hence substitute x = 1 , y = 2

2(1) - 2 + a = 0

=> 2 - 2 +  a = 0

=> a = 0

Value of a is 0

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