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Answer 1 · 2016-09-01T17:25:36

$x^{2} + y^{2} - 6 x + 2 y - 15 = 0$ $x^2+y^2-6x+2y-15=0$

Let $l_{1} : 4 x - 3 y + 10 = 0, and, l_{2} : 3 x + 4 y - 30 = 0$ $l_1 : 4x-3y+10=0, and, l_2 : 3x+4y-30=0$ be the

given tgt. lines to the reqd. circle, say, $S$ $S$ . Note taht, $l_{1} ⊥ l_{2}$ $l_1 bot l_2$ .

Also, let $P (- 1, 2) and Q (6, 3)$ $P(-1,2) and Q(6,3)$ be the pts. of contact of $S$ $S$ with

$l_{1} and l_{2}$ $l_1 and l_2$ , resp.

Let pt. $C$ $C$ be the Centre, and, $r$ $r$ the radius, of $S$ $S$ . Then, we

know, from Geometry, that,

$dist. C P = dist. C Q = r, and, line C P ⊥ l_{1}, line C Q ⊥ l_{2}$ $"dist."CP="dist."CQ =r, and, "line "CP bot l_1, "line "CQ bot l_2$ .

Since, $line CP \cap line C Q = {C}$ $"line CP"nn"line "CQ={C}$ , we will obtain the co-ords. of

the centre $C$ $C$ by solving the eqns. of these lines.

Eqn. of line CP :-

$Line C P ⊥ l_{1}, &, l_{2} ⊥ l_{1}$ $"Line "CP botl_1, &, l_2botl_1$

$\Rightarrow line C P ∣ ∣ l_{2} : 3 x + 4 y - 30 = 0, with P \in C P$ $rArr " line "CP |\| l_2 :3x+4y-30=0," with "P in CP$

$" Eqn. of "CP : 3x+4y=3(-1)+4(2)=5.........(1)$

Similarly, Eqn. of line CQ $: 4x-3y=15..................(2)$

Solving $(1), &, (2)$ , we get, the Centre $C(3,-1)$ .

Then, $r=dist. CP=sqrt{(3+1)^2+(2+1)^2}=5$

Finally, the eqn. of $S : (x-3)^2+(y+1)^2=5^2$ , or,

$x^2+y^2-6x+2y-15=0$

Enjoy Maths.!