Question

.How do I find the value of T, if the straightline y= T is the tangent to locus P with an equation of x^2 + y^2-2x+5y+5 =0?
​

Answer 1

Equation of circle =x

2

+y

2

−22x−4y+25=0

(x−11)

2

+(y−2)

2

=10

2

, radius 10 and r=(11,2)

Slope of line 5x+12y+8=0 is m

1

=

dx

dy

=

12

−5

So as we know two perpendicular has product of slope 2−1 i.e. m

1

m

2

2−1 where m

1

and m

2

are slope of two ⊥ lines

Here, if m

1

=

12

−5

then m

2

=

5

12

General equation of line with slope

5

12

⇒

y=

5

12

x+c.....(1)

perpendicular distance of tangent from center =10 (radius)

So,

(

5

12

)

2

+1

∣

∣

∣

∣

∣

5

12

(11)+C−2

∣

∣

∣

∣

∣

=10 as ⊥ distance =

a

2

+b

2

∣ax

1

+by

1

+c∣

from pt

∣5c+122∣=130

5c+122=130 5c+122=−130

C=

5

8

C=

5

−252

Put value of c in (1)

5y=12x+8 and 5y=12x−252 are lines ⊥ to 5x+12y+8=0 and are tangent to given will

Answer 2

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