(2)

One side of square is y = x + k

Distance of (5, 3) to the line y = x + k is

$\frac{|3 – 5 – k|}{\sqrt{2}} = \sqrt{2}$

$\Rightarrow |–2 – k| = 2 \Rightarrow k = 0 \mathrm{or} k = –4$

So lines are y = x and y = x – 4

Now, solving these lines with circle

y = x and $x^2 + y^2 – 10x – 6y + 30 = 0$

$\Rightarrow 2x^2 – 16x + 30 = 0$

$\Rightarrow$    x = 3, y = 3 and x = 5, y = 5

and y = x – 4 and $x^2 + y^2 – 10x – 6y + 30 = 0$

$\Rightarrow$    x = 5, y = 1 and x = 7, y = 3

$\sum _{i = 1}^4 x_i^2 + y_i^2$ = 9 + 9 + 25 + 25 + 25 + 1 + 49 + 9 = 152.

(1)

Distance of centre from chord $PQ =$$\left|\frac{0 + 0 –2}{\sqrt{1^2 + 1^2}}\right|$$= \frac{2}{\sqrt{2}} = \sqrt{2}$

Let distance of centre from chord MN = p, then length of chord = $2\sqrt{10 – p^2}$

$\Rightarrow 2\sqrt{10 – p^2} = 2$            ($\because$   Given)

$\Rightarrow 10 – p^2 = 1 \Rightarrow p^2 = 9 \Rightarrow p = 3$

$\Rightarrow$    Distance between chord PQ and MN = $3 – \sqrt{2}$.

(3)

Centre of circle = (2, 1)

$\therefore$   Equation of circle will be

${(x – 2)}^2 + {(y – 1)}^2 = 1$

Distance between C(2, 1) and P(5, 5)

= $\sqrt{{(2 – 5)}^2 + {(1 – 5)}^2}$

= 5 units

Also, AC = BC = 1 unit               [$\because$   Radius = 1]

$\therefore$   Shortest distance of circle from P = 5 – 1 = 4 units.

(1)

We have,  $C_1 : x^2 + y^2 – 2(x + y) + 1 = 0$

                 $C_2 : {(x + 1)}^2 + y^2 – 4 = 0$

For common chord, we have $C_1 – C_2 = 0$

$\Rightarrow x^2 + y^2 –2x – 2y + 1 – x^2 – y^2 – 2x + 3 = 0$

$\Rightarrow –4x – 2y + 4 = 0$

$\Rightarrow 2x + y – 2 = 0$

Since, common chord intersects y-axis

So,  x = 0

$\Rightarrow$  y = 2

So, point of intersection of common chord with y-axis is P(0, 2).

Required distance = ${\left(\sqrt{{(1 – 0)}^2 + {(1 – 2)}^2}\right)}^2 = 2$.

(2)

Let r be the radius of circle and centre at C(r, r).

Then, $F{C}^2 = r^2$

$\Rightarrow {(r –2)}^2 + {(r – 2)}^2 = r^2$

$\Rightarrow r^2 – 8r + 8 = 0$.

(4)

$B(\alpha , \beta ), C(8,\frac{15}{2})$ are the centres of circle $C_1$ and $C_2$ respectively. Now, A(6, 6) is the given point.

A divides BC in ratio 2 : 1

$\Rightarrow \frac{16+\alpha }{3}=6$  and  $\frac{15+\beta }{3}=6$

$\Rightarrow \alpha =2 \mathrm{and} \beta =3$

Also, $\overline{BC}=r_1+r_2 \Rightarrow r_1+r_2=\sqrt{{(2–8)}^2+{(3–\frac{15}{2})}^2}$

$\Rightarrow 2r_2+r_2=\sqrt{36+\frac{81}{4}}$              $[\because r_1 : r_2=2 : 1]$

$\Rightarrow 3r_2=\frac{15}{2} \Rightarrow r_2=\frac{5}{2} \mathrm{units} \Rightarrow r_1=5 \mathrm{units}$

$\therefore (\alpha +\beta )+4(r_1^2+r_2^2)=5+4(25+\frac{25}{4})=130$.

(4)

$\frac{x+4}{1}=\frac{y–5}{2}=\frac{–2(–4+2\times 5–2)}{1+4}=\frac{–8}{5}$

$\Rightarrow x=–4–\frac{8}{5}= –\frac{28}{5}, y=5–\frac{16}{5}=\frac{9}{5}$

$\therefore \mathrm{Image} \mathrm{is} (\frac{–28}{5},\frac{9}{5})$

Image lies on circle ${(x+4)}^2+{(y–3)}^2=r^2$

${(\frac{–28}{5}+4)}^2+{(\frac{9}{5}–3)}^2=r^2$

$\Rightarrow \frac{64}{25}+\frac{36}{25}=r^2 \Rightarrow r^2=4 \Rightarrow r=2$

(3)

We have, 3x + 5y = 1          ... (i)

  $(2+c)x+5c^{2}y=1$       ... (ii)

Multiplying (i) by $c^2$ and subtracting it from (ii), we get

$(2+c–3c^2)x=1–c^2$

$\Rightarrow x_c=\frac{1–c^2}{2+c–3c^2}=\frac{(1–c)(1+c)}{(1–c)(3c+2)}=\frac{c+1}{3c+2}$

$\therefore y=\frac{1–3x}{5}=\frac{1–3\left({\displaystyle \frac{c+1}{3c+2}}\right)}{5}=\frac{3c+2–3c–3}{5(3c+2)}$

$y_c=\frac{–1}{5(3c+2)}$

Now, $h=\underset{c\to 1}{\lim }{x}_c=\underset{c\to 1}{\lim }\left(\frac{c+1}{3c+2}\right)=\frac{1+1}{3+2}=\frac{2}{5}$

and $k=\underset{c\to 1}{\lim }{y}_c=\underset{c\to 1}{\lim }\frac{–1}{5(3c+2)}=\frac{–1}{25}$

Equation of circle is

${(x–\frac{2}{5})}^2+{(y+\frac{1}{25})}^2=\frac{64}{25}+\frac{1}{625}$              [$\because$ Circle passes through (2, 0)]

$\Rightarrow x^2+y^2–\frac{4}{5}x+\frac{2}{25}y+\frac{4}{25}+\frac{1}{625}=\frac{64}{25}+\frac{1}{625}$

$\Rightarrow 25x^2+25y^2–20x+2y+4=64$

$\Rightarrow 25x^2+25y^2–20x+2y–60=0$

(2)

We have, $C : x^2+y^2=4$                         ... (i)

and $C\text{'} : x^2+y^2–4\lambda x+9=0$

$\Rightarrow C\text{'} : {(x–2\lambda )}^2+y^2=4{\lambda }^2–9$           ... (ii)

Radius of C, $r_1=2$ and radius of C', $r_2=\sqrt{4{\lambda }^2–9}$

When two circles intersect at two points,

then $|r_1–r_2|<CC\text{'}<|r_1+r_2|$

$\Rightarrow$$|2–\sqrt{4{\lambda }^2–9}|$$<2\lambda <2+\sqrt{4{\lambda }^2–9}$              ...(iii)

By (iii), we have $4+4{\lambda }^2–9–4\sqrt{4{\lambda }^2–9}<4{\lambda }^2$

$\Rightarrow –5–4\sqrt{4{\lambda }^2–9}<0$

$\Rightarrow 5+4\sqrt{4{\lambda }^2–9}>0 \Rightarrow \sqrt{4{\lambda }^2–9}>–\frac{5}{4}$

On Squaring both sides, we get

$4{\lambda }^2–9>\frac{25}{16} \Rightarrow 4{\lambda }^2>\frac{169}{16} \Rightarrow {\lambda }^2>\frac{169}{64}$

$\Rightarrow \lambda >\frac{13}{8} \mathrm{or} \lambda <–\frac{13}{8}$

$\therefore \lambda \in (–\infty ,–\frac{13}{8})\cup (\frac{13}{8},\infty )$

Also, $4{\lambda }^2<{(2+\sqrt{4{\lambda }^2–9})}^2$           [By (iii)]

$\Rightarrow 4{\lambda }^2<4+4\sqrt{4{\lambda }^2–9}+4{\lambda }^2–9$

$\Rightarrow 0<–5+4\sqrt{4{\lambda }^2–9} \Rightarrow 5<4\sqrt{4{\lambda }^2–9}$

On squaring, we get

$\frac{25}{16}<4{\lambda }^2–9 \Rightarrow \frac{169}{64}<{\lambda }^2$

$\therefore \lambda \in (–\infty ,–\frac{13}{8})\cup (\frac{13}{8},\infty )$

Thus, Circles C and C' intersect at two distinct points for

$\lambda \in R–[\frac{–13}{8},\frac{13}{8}]$

$\therefore a=\frac{–13}{8}, b=\frac{13}{8}$

$\therefore$  (8a + 12, 16b –20) = (–1, 6) which satisfies only $6x^2+y^2= 42$.

(3)

Let $(x_1,y_1)$ be the mid point of chords.

So, equation of chord of the circle $x^2+y^2– 2y=0$ is,

$x·x_1+y·y_1–(y+y_1)=x_1^2+y_1^2–2y_1$

The chord is passing through origin

$\therefore –y_1=x_1^2+y_1^2–2y_1$

        $x_1^2+y_1^2–y_1=0$               ... (i)

Now, (i) intersects the line x + y = 1

$\therefore {(1–y_1)}^2+y_1^2–y_1=0$

$\Rightarrow 1+y_1^2–2y_1+y_1^2–y_1=0 \Rightarrow 2y_1^2–3y_1+1=0$

$\Rightarrow (2y_1–1)(y_1–1)=0 \Rightarrow y_1=\frac{1}{2} \mathrm{or} y_1=1$

If $y_1=\frac{1}{2}$, then $x_1=\frac{1}{2}$                [From (i)]

If y = 1, then $x_1=0$

So, $P=(\frac{1}{2},\frac{1}{2})$ and Q = (1, 0)   $\therefore PQ=\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{–1}{2}\right)}^2}=\frac{1}{\sqrt{2}}$