Explanation
Given:
The equation of line is $$ x - 2y = 3 $$
Substitute $$x = 0$$ in the given equation.
$$ 0 - 2y = 3 \Rightarrow - 2y = 3 \Rightarrow y = - \dfrac{3}{2}$$
Substitute $$y = 0$$ in the given equation.
$$ x - 2\left (0 \right ) = 3 \Rightarrow x = 3 $$
The points are $$ \left (3, 0 \right ) \, and \, \left (0, - \dfrac{3}{2} \right ) $$.
Let the centre be $$ \left (a, -a \right ) $$.
Substitute $$ \left (a, -a \right ) $$ in given equation.
$$ a + 2a = 3 \Rightarrow 3a = 3 \Rightarrow a = 1 $$
Centre $$ = \left (1, -1 \right ) $$
Radius $$ = 1$$
The equation of a circle becomes $$ \left (x - 1 \right )^{2} + \left (y + 1 \right )^{2} = 1 $$.
Hence, the required solution is found.
In the $$xy$$ plane, the segment with end points$$(3,8)$$ and $$(-5,2)$$ is the diameter of the circle. The point $$(k,10)$$ lies on the circle for:
$$C\left( x,y \right)=\left( \dfrac{3+5}{2},\dfrac{8+2}{2} \right)=\left( 4,5 \right)$$
The length of the diameter of the circle is,
$$D=\sqrt{{{\left( 5-3 \right)}^{2}}+{{\left( 2-8 \right)}^{2}}}=2\sqrt{10}\ units$$
Therefore,
Radius $$=\sqrt{10}\ units$$
Therefore, the equation of the required circle is,
$${{\left( x-4 \right)}^{2}}+{{\left( y-5 \right)}^{2}}=10$$
Since, the point $$\left( k,10 \right)$$ lies on the circle, we have
$$ {{\left( k-4 \right)}^{2}}+{{\left( 10-5 \right)}^{2}}=10 $$
$$ {{k}^{2}}-8k+16+25=10 $$
$$ {{k}^{2}}-8k+31=0 $$
$$ 2x^{2} + 3y - 8x - 18y + 35 = \lambda $$
$$ 2\left (x^{2} - 4x \right ) + 3 \left ( y^{2} - 6y + 35 \right ) = \lambda $$
$$ 2\left (x - 2 \right )^{2} + 3 \left ( y - 3 \right )^{2} = \lambda $$
For $$ \lambda = 0 $$, then
$$ 2\left (x - 2 \right )^{2} + 3 \left ( y - 3 \right )^{2} = 0 $$
Thus, the point is $$ \left ( 2,3 \right ) $$.
Hence, the correct option is ‘d’.
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