Question

Q.11 Find the lengths of the intercepts made on
the co-ordinate axes, by the circle.
(i) x2 + y2 – 8x + y - 20 = 0​

Answer 1

Answer:

Let the equation of any circle be

x

2

+y

2

+2gx+2fy+c=0(i)

For intercept made by the circle on x-axis, put y=0 in(i)

⇒x

2

+2gx+c=0

If x

1

,x

2

are roots of (ii), then length of the intercept on x-axis is

∣x

1

−x

2

∣=

(x

1

+x

2

)

2

−4x

1

x

2

=2

g

2

−c

Similarly length of the intercept of the yaxis is 2

f

2

−c

Since the lengths of these intercepts are equal

g

2

−c

=

f

2

−c

⇒g

2

=f

2

=(−g)

2

=(−f)

2

Therefore, centre lies on x

2

−y

2

=0

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Answer 2

Let the equation of any circle be

x

2

+y

2

+2gx+2fy+c=0(i)

For intercept made by the circle on x-axis, put y=0 in(i)

⇒x

2

+2gx+c=0

If x

1

,x

2

are roots of (ii), then length of the intercept on x-axis is

∣x

1

−x

2

∣=

(x

1

+x

2

)

2

−4x

1

x

2

=2

g

2

−c

Similarly length of the intercept of the yaxis is 2

f

2

−c

Since the lengths of these intercepts are equal

g

2

−c

=

f

2

−c

⇒g

2

=f

2

=(−g)

2

=(−f)

2

Therefore, centre lies on x

2

−y

2

=0