Two circles , each of radius 5, have a common tangent at (1,1) whose equation is 3x + 4y - 7 = 0. Then their centres are

a. (4,-5), (-2,3)

b. (4,-3), (-2,5)

c. (4,5), (-2,-3)

d. n.o.t.

Question

Two circles , each of radius 5, have a common tangent at (1,1)
whose equation is 3x + 4y - 7 = 0 Then their centres are
a (4,-5), (-2,3)
b (4,-3), (-2,5)
c (4,5), (-2,-3)
d n o t

Pronay · Accepted Answer

equation of tangent is 3x + 4y - 7 = 0 ------------------(I)

radius = 5

let centre of circle be (h, k )

distance of centre from the tangent is equal to radius of circle

so $| \frac{3 h + 4 k - 7}{{(3^{2} + 4^{2})}^{\frac{1}{2}}} | = 5$

or 3h+4k - 7= 25
or 3h+4k = 32 or h = (32-4k)/3------------------------------------(II)

equation of circle whose centre is ( h, k) & radius 5 is

${(x - h)}^{2} + {(y - k)}^{2} = 25$
this circle passise through the point ( 1,2 )
${(1 - h)}^{2} + {(1 - k)}^{2} = 25 h^{2} - 2 h + 1 + k^{2} - 2 k + 1 = 25 h^{2} - 2 h + k^{2} - 2 k - 23 = 0$ ----------------(III)
from(II) & (III)
${(\frac{32 - 4 k}{3})}^{2} - 2 \times \frac{(32 - 4 k)}{3} + k^{2} - 2 k - 23 = 0$

$\frac{1024 + 16 k^{2} - 256 k}{9} - \frac{64 - 8 k}{3} + k^{2} - 2 k - 23 = 0 \frac{{16 k}^{2} - 256 k + 1024 - 192 + 24 k + 9 k^{2} - 18 k - 207}{9} = 0 25 k^{2} - 250 k + 625 = 0 k^{2} - 10 k + 25 = 0 k^{2} - 5 k - 5 k + 25 = 0 (k - 5) (k - 5) = 0 k = 5 s o, h = \frac{32 - 5 \times 4}{3} = \frac{12}{3} = 4$

so centre is ( 5,5 )

Two circles , each of radius 5, have a common tangent at (1,1) whose equation is 3x + 4y - 7 = 0. Then their centres are a. (4,-5), (-2,3) b. (4,-3), (-2,5) c. (4,5), (-2,-3) d. n.o.t.

Two circles , each of radius 5, have a common tangent at (1,1) whose equation is 3x + 4y - 7 = 0. Then their centres are

a. (4,-5), (-2,3)

b. (4,-3), (-2,5)

c. (4,5), (-2,-3)

d. n.o.t.