Explanation
Given that,
Equations of diameter $$3x-4y-7=0$$ …… (1) and $$2x-3y-5=0$$…… (2)
Solve equations are
$$3x-4y-7=0$$ $$×2$$
$$2x-3y-5=0$$ $$×3$$
$$ 6x-8y-14=0 $$
$$ \underline{6x-9y-15=0}\,\,\,\,\,\,\,\,on\,\,subtracting\,\,\,that $$
$$ y+1=0 $$
$$ y=-1 $$
Put the value of in equation (1)
$$3x-4y-7=0$$
$$ 3x-4\left( -1 \right)-7=0 $$
$$ 3x+4-7=0 $$
$$ 3x-3=0 $$
$$ x=1 $$
Hence , the coordinates of the centre is $$(1,-1)$$
Also given area of circle= $$49\pi$$
$$ \pi {{r}^{2}}=49\pi $$
$$ {{r}^{2}}=49 $$
$$ r=7 $$
Then, we know that the equation of circle is
$$ {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}} $$
$$ {{\left( x-1 \right)}^{2}}+{{\left( y+1 \right)}^{2}}={{7}^{2}} $$
$$ {{x}^{2}}+1-2x+{{y}^{2}}+1+2y=49 $$
$$ {{x}^{2}}+{{y}^{2}}-2x+2y+2=49 $$
$$ {{x}^{2}}+{{y}^{2}}-2x+2y=49-2 $$
$$ {{x}^{2}}+{{y}^{2}}-2x+2y=47 $$
$$ {{x}^{2}}+{{y}^{2}}-2x+2y-47=0 $$
Hence, it is complete solution.
Option $$(C)$$ is correct answer.
We know that equation of ellipse is
$$ \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 $$ …….(1)
Given that
$$ e=\sqrt{\dfrac{2}{5}} $$
$$ \sqrt{\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}}}=\sqrt{\dfrac{2}{5}} $$
$$ 5{{a}^{2}}-5{{b}^{2}}=2{{a}^{2}} $$
$$ {{a}^{2}}=\dfrac{5{{b}^{2}}}{3} $$ …….(2)
$$\because $$ ellipse pass through (-3,1)
Then $$x=-3, y=1$$
Put in equation (1) we get
$$ \dfrac{{{\left( -3 \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{1}^{2}}}{{{b}^{2}}}=1 $$
$$ {{a}^{2}}+9{{b}^{2}}={{a}^{2}}{{b}^{2}} $$
$$ \dfrac{5{{b}^{2}}}{3}+9{{b}^{2}}=\dfrac{5{{b}^{2}}}{3}.{{b}^{2}} $$
$$ {{b}^{2}}=\dfrac{32}{5} $$ (From equation (1) and (2) )
Put in equation (2) , we get $${{a}^{2}}=\dfrac{32}{3}$$
the value of a and b put in equation (1), we get
$$ \dfrac{{{x}^{2}}}{\dfrac{32}{3}}+\dfrac{{{y}^{2}}}{\dfrac{32}{5}}=1 $$
$$ 3{{x}^{2}}+5{{y}^{2}}=32 $$
This is required equation
$$\textbf{Hence, Option 'B' is correct} $$
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