Question

xty
find the Equations of the circles which touch
sy +=
Qx-3y +
at and having oradius vis.​

Answer 1

Answer:

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Find the equation of circles which touch 2x−3y+1=0 at (1,1) and having radius

13

Medium

Solution

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Let the equation of the circle.

(x−h)

2

+(y−k)

2

=r

2

----- (1)

(h,k) is the coordinate of center, radius

13

and it is touches by the line 2x−3y+1=0 at (1,1)

By distance formula

(h−1)

2

−(k−1)

2

=

13

h

2

=13−k

2

+2k+2h−2 ------- (2)

Now,

Tangent is always perpendicular to the radius

Slope of the line 2x−3y+1(m

1

)=

3

2

Slope of the radius m

2

=−

2

3

Also the slope of of radius

m

2

=

h−1

k−1

=−

2

3

3h+2k=−5 ------- (3)

h=

3

2k−5

----- (4)

squaring on both side on equation (3)

9h

2

+4k

2

=25

From (2) and (4)

5k

2

−6k−104

(5k−26)(k+4)=0

k=−4,

5

26

Therefore,

h=−

3

13

,

15

27

Hence the point of center

(−4,−

3

13

),(

5

26

,

15

27

)

Hence the equation of circle are,

(x−(−4))

2

+(y−(−

3

13

))

2

=(

13

)

2

or

(x−

5

26

)

2

+(y−

5

27

)

2

=(

13

)

2