Question

the normal of the circle(x-2)*2+(y-1)*2=16 which bisects the chord cut off by the line x-2y+3=0

Answer 1

A bonus point for normal of the circle is , it always passes through centre of circle. here equation of circle is (x - 2)² + (y - 1)² = 16.

so, centre = (2, 1)

let m is slope of normal. now equation of normal passes through (2, 1) is ...

y - 1 = m(x - 2)

y = mx - 2m + 1 .....(1)

a/c to question,

normal bisects the chord cut off by the line x - 2y + 3 = 0.

but we know, a line passing through centre , bisects the chord perpendicularly.

so, normal and chord are perpendicular to each other.

so, slope of normal × slope of chord = -1

or, m × 1/2 = -1

or, m = -2 , putting it in equation (1)

so, y = -2x +2 + 1

or, 2x + y - 3 = 0

hence, equation of normal of the circle is 2x + y - 3 = 0

Answer 2

Answer:

2x + y - 5 = 0

Step-by-step explanation:

A bonus point for normal of the circle is , it always passes through centre of circle. here equation of circle is (x - 2)² + (y - 1)² = 16.

so, centre = (2, 1)

let m is slope of normal. now equation of normal passes through (2, 1) is ...

y - 1 = m(x - 2)

y = mx - 2m + 1 .....(1)

a/c to question,

normal bisects the chord cut off by the line x - 2y + 3 = 0.

but we know, a line passing through centre , bisects the chord perpendicularly.

so, normal and chord are perpendicular to each other.

so, slope of normal × slope of chord = -1

or, m × 1/2 = -1

or, m = -2

so substituting m in y = mx - 2m + 1
y = (-2)x - 2(-2) + 1

y = -2x + 4 + 1

2x + y - 5 = 0

Equation of normal of the circle is 2x + y - 5 = 0.
Hence Proved.