Consider three circles: C_1:x^2 + y^2 = r^2 C_2:(x - 1)^2 + (y - 1)^2 = r^2 C_3:(x - 2)^2 …

Question

Consider three circles:

${C_1}:{x^2} + {y^2} = {r^2}$

${C_2}:{(x - 1)^2} + {(y - 1)^2} = {r^2}$

${C_3}:{(x - 2)^2} + {(y - 1)^2} = {r^2}$

If a line L : y = mx + c be a common tangent to C1, C2 and C3 such that C1 and C3 lie on one side of line L while C2 lies on other side, then the value of $20({r^2} + c)$ is equal to :

Accepted Answer

${c_1}:{x^2} + {y^2} = {r^2}$ ; center = (0, 0) and radius = r

${c_2}:{(x - 1)^2} + {(y - 1)^2} = {r^2}$ ; center = (1, 1) and radius = r

${c_3}:{(x - 2)^2} + {(y - 1)^2} = {r^2}$ ; center = (2, 1) and radius = r

Distance of $y = mx + c$ line from center (0, 0) is,

$$\left| {{{0 + 0 + c} \over {\sqrt {{m^2} + 1} }}} \right| = r$$ ..... (1)

Distance of $y = mx + c$ line from center (1, 1) is,

$$\left| {{{m - 1 + c} \over {\sqrt {{m^2} + 1} }}} \right| = r$$ ..... (2)

Distance of $y = mx + c$ line from center (2, 1) is,

$$\left| {{{2m - 1 + c} \over {\sqrt {{m^2} + 1} }}} \right| = r$$ .... (3)

From (1) and (2), we get

$$\left| {{c \over {\sqrt {1 + {m^2}} }}} \right| = \left| {{{m - 1 + c} \over {\sqrt {1 + {m^2}} }}} \right|$$

$$ \Rightarrow m - 1 + c =  \pm \,c$$ ..... (4)

taking positive sign,

$m - 1 + c = c$

$$ \Rightarrow m - 1 = 0$$

$$ \Rightarrow m = 1$$

From (2) and (3), we get

$$\left| {{{m - 1 + c} \over {\sqrt {1 + {m^2}} }}} \right| = \left| {{{2m - 1 + c} \over {\sqrt {{m^2} + 1} }}} \right|$$

$$ \Rightarrow (m - 1 + c) = \, \pm \,(2m - 1 + c)$$ ...... (5)

taking positive sign,

$m - 1 + c = 2m - 1 + c$

$$ \Rightarrow m = 0$$

By taking positive sign we get two different value of m so it is not acceptable.

From equation (4), taking negative sign,

$m - 1 + c =  - c$

$$ \Rightarrow m + 2c - 1 = 0$$ ..... (6)

From equation (5), taking negative sign

$m - 1 + c =  - (2m - 1 + c)$

$$ \Rightarrow 3m + 2c - 2 = 0$$ ..... (7)

Solving equation (6) and (7), we get

$3m + 1 - m - 2 = 0$

$$ \Rightarrow 2m = 1$$

$$ \Rightarrow m = {1 \over 2}$$

$$\therefore$$ $$2c = 1 - {1 \over 2}$$

$$ \Rightarrow c = {1 \over 4}$$

Putting value of $$m = {1 \over 2}$$ and $$c = {1 \over 4}$$ in equation (1), we get

$$r = \left| {{{{1 \over 4}} \over {\sqrt {1 + {1 \over 4}} }}} \right|$$

$$ = \left| {{1 \over 4} \times {2 \over {\sqrt 5 }}} \right|$$

$$ = {1 \over {2\sqrt 5 }}$$

$$\therefore$$ $20({r^2} + c)$

$$ = 20\left( {{1 \over {4 \times 5}} + {1 \over 4}} \right)$$

$$ = 20\left( {{{1 + 5} \over {20}}} \right)$$

$= 6$

Solutions · 1