Given, x $$\ge$$ 0, y² $$\le$$ 4 $$-$$ x

Let equation of circle be

(x $$-$$ h)² + y² = h² .... (i)


Solving Eq. (i) with y² = 4 $$-$$ x, we get

x² $$-$$ 2hx + 4 $$-$$ x = 0

$$\Rightarrow$$ x² $$-$$ x(2h + 1) + 4 = 0 .... (ii)

For touching/tangency, Discriminant (D) = 0

i.e. (2h + 1)² = 16 $$\Rightarrow$$ 2h + 1 = $$\pm$$ 4

$$\Rightarrow$$ 2h = $$\pm$$ 4 $$-$$ 1

$$\Rightarrow$$ $$h = {3 \over 2},h = {{ - 5} \over 2}$$ (Rejected) because part of circle lies outside R. So, $$h = {3 \over 2}$$ = radius of circle (c).

Putting h = 3/2 in Eq. (ii),

x² $$-$$ 4x + 4 = 0 $$\Rightarrow$$ (x $$-$$ 2)² = 0 $$\Rightarrow$$ x = 2

So, $$\alpha$$ = 2

Given, x $$\ge$$ 0, y² $$\le$$ 4 $$-$$ x

Let equation of circle be

(x $$-$$ h)² + y² = h² .... (i)


Solving Eq. (i) with y² = 4 $$-$$ x, we get

x² $$-$$ 2hx + 4 $$-$$ x = 0

$$\Rightarrow$$ x² $$-$$ x(2h + 1) + 4 = 0 .... (ii)

For touching/tangency, Discriminant (D) = 0

i.e. (2h + 1)² = 16 $$\Rightarrow$$ 2h + 1 = $$\pm$$ 4

$$\Rightarrow$$ 2h = $$\pm$$ 4 $$-$$ 1

$$\Rightarrow$$ $$h = {3 \over 2},h = {{ - 5} \over 2}$$ (Rejected) because part of circle lies outside R. So, $$h = {3 \over 2}$$ = radius of circle (C).

As we know that the equation of family of circles passes through the points of intersection of given circle x² + y² = r² and line PQ : 2x + 4y = 5 is,

(x² + y² $$-$$ r²) + $$\lambda $$(2x + 4y $$-$$ 5) = 0 ......(i)

Since, the circle (i) passes through the centre of circle

x² + y² = r²,

So, $$-$$ r² $$-$$ 5$$\lambda $$ = 0

or 5$$\lambda $$ + r² = 0 ....(ii)

and the centre of circle (i) lies on the line x + 2y = 4, so centre ($$-$$ $$\lambda $$, $$-$$ 2$$\lambda $$) satisfy the line x + 2y = 4.

Therefore, $$-$$$$\lambda $$ $$-$$4$$\lambda $$ = 4

$$ \Rightarrow $$ $$-$$5$$\lambda $$ = 4

$$ \Rightarrow $$ r² = 4 {from Eq. (ii)}

$$ \Rightarrow $$ r = 2

According to given information the figure is as following



From the figure,

$$AC = {2 \over {\sin \theta }}$$ ....(i)

$$ \because $$ $$\sin \theta = {1 \over {CB}}$$ (from $$\Delta $$CPB) .... (ii)

and $$\sin \theta = {2 \over {AC}} = {2 \over {CB + AB}}$$ (from $$\Delta $$CQA) ....(iii)

$$ \because $$ AB = AM + MB = 2AM [$$ \because $$ AM = MB]

$$ = 2{{\left| {(8 \times 2) - (6 \times 3) - 23} \right|} \over {\sqrt {64 + 36} }}$$

$$ = {{2 \times 25} \over {10}} = 5.00$$

From Eqs. (ii) and (iii), we get

$$\sin \theta = {1 \over {CB}} = {2 \over {CB + AB}}$$

$$ \Rightarrow {1 \over {CB}} = {2 \over {CB + 5}}$$ [$$ \because $$ AB = 5]

$$ \Rightarrow CB + 5 = 2CB \Rightarrow CB = 5 = {1 \over {\sin \theta }}$$

From the Eq. (i), we get

$$AC = {2 \over {\sin \theta }} = 2 \times 5 = 10.00$$