Question

The equation of the tangent to the circle at A(1,8) is 3x-2y+5=0 and its centre is on the line 7x+3y-1=0 . Find the centre of the circle

Answer 1

Step-by-step explanation:

Use the standard form to write two equations using points A and B:

(

−

2

−

h

)

2

+

(

0

−

k

)

2

=

r

2

(

5

−

h

)

2

+

(

1

−

k

)

2

=

r

2

Because

r

2

=

r

2

, we can set the left sides equal:

(

−

2

−

h

)

2

+

(

0

−

k

)

2

=

(

5

−

h

)

2

+

(

1

−

k

)

2

Expand the squares using the pattern

(

a

−

b

)

2

=

a

2

−

2

a

b

+

b

2

4

+

4

h

+

h

2

+

k

2

=

25

−

10

h

+

h

2

+

1

−

2

k

+

k

2

Combine like terms (noting that the squares cancel):

4

+

4

h

=

25

−

10

h

+

1

−

2

k

Move the k term the left and all other terms to the right:

2

k

=

−

14

h

+

22

Divide by 2

k

=

−

7

h

+

11

[1]

Evaluate the given line at the center point:

2

h

+

k

−

1

=

0

Write in slope-intercept form

k

=

−

2

h

+

1

[2]

Subtract equation [2] from equation [1]:

k

−

k

=

−

7

h

+

2

h

+

11

−

1

0

=

−

5

h

+

10

h

=

2

Substitute 2 for h in equation [2]

k

=

−

2

(

2

)

+

1

k

=

−

3

Substitute the center

(

2

,

−

3

)

into the equation of a circle using point A and solve for the value of r:

(

−

2

−

2

)

2

+

(

0

−

−

3

)

2

=

r

2

(

−

4

)

2

+

3

2

=

r

2

r

2

=

25

r

=

5

Substitute the center

(

2

,

−

3

)

and #r = 5 into the general equation of a circle, to obtain the specific equation for this circle:

(

x

−

2

)

2

+

(

y

−

−

3

)

2

=

5

2