Question

1. Find 'k' if the following pairs of circles are orthogonal x2 + y2 - 5x - 14y - 3k = 0 and
x2 + y2 + 2x+4y+k=0​

Answer 1

Step-by-step explanation:

Equation of circle-

S

1

:x

2

+y

2

−5x−14y−34=0⇒(x−

2

5

)

2

+(y−7)

2

−

4

25

−49−34=0

⇒S

1

:(x−

2

5

)

2

+(y−7)

2

=(

2

357

)

2

Here r

1

=

2

357

C

1

=(

2

5

,7)

S

2

x

2

+y

2

+2x+4y+k=0⇒(x+1)

2

+(y+2)

2

−1−4+k=0

⇒S

1

:(x+1)

2

+(y+2)

2

=(

5−k

)

2

Here r

1

=

5−k

C

1

=(−1,−2)

: C

1

C

2

=

(−1−

2

5

)

2

+(−2−7)

2

=

2

373

As we know that the angle between the two circles is given as-

cosθ=

r

1

r

2

r

1

2

+r

2

2

−d

2

Here d is the distance between centre of circle.

Since the circles orthogonal, i.e., θ=

2

π

∴cos

2

π

=

r

1

r

2

(

2

357

)

2

+(

5−k

)

2

−(

2

373

)

2

4

357

+5−k−

4

373

=0

357+20−4k−373=0

k=0

Hence the value of k for which the circles are orthogonal is 0.