Question

Find the equation of the chord of the circle x^2+y^2-4x=0 which is bisected at the point (1 1) brainly

Answer 1

Answer:

Equation of the required chord x-y= 0

Step-by-step explanation:

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Answer 2

Answer:

Equation of Chord is x = y

Step-by-step explanation:

Given:

Equation of circle, x² + y² - ax = 0

Mid point of chord = ( 1 , 1 )

To find: Equation of chord.

We know that perpendicular from center of the circle bisect the chord.

So,

if m is slope of chord and n is slop of perpendicular line from center to mid point of chord, then use result of slope of perpendicular lines.

we have m × n = -1 ....................(1)

Equation of circle,

x² + y² - 4x = 0

x² - 4x + y² = 0

x² - 4x + 2² + y² = 0 + 2²

( x - 2 )² + y² = 4

⇒ Center of the circle = ( 2 , 0 )

Slope of the line from center to mid point of chord, n = $\frac{y_2-y_1}{x_2-x_1}=\frac{1-0}{1-2}=\frac{1}{-1}=-1$

From (1),

m × (-1) = -1

m = 1

Now using slope-point form we have,

$(y-y_1)=m(x-x_1)$

$(y-1)=1(x-1)$

y - 1 = x - 1

x - y = 0

Therefore, Equation of Chord is x = y