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# Find the equation of the family. Of circles which touch the pair of straight lines x^2-y^2+2y-1=0

## 1 Answers

3 years ago
Dear Student,

First lets get both the straight lines

Solve for y:
-1 + x^2 + 2 y - y^2 = 0
Multiplying both sides by -1:
1 - x^2 - 2 y + y^2 = 0
y = (2 ± sqrt(4 - 4 (1 - x^2)))/(2) = 1 ± sqrt(x^2):
so they are y=1+x and y=1-x

now they both are tangents to (x-h)^2+(y-k)^2 = 1/2

condition for tangency: c=±a*sqrt(1+m^2)

=>1=±a*sqrt(1+1)

=>a=±1/sqrt(2)

so eqn is (x-h)^2 +(y-k)^2 = a^2.

Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

or, (x-h)^2 +(y-k)^2 = 1/2   [Ans]

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