TG EAMCET 2 May 2025 Shift 1 Paper

The equation of the circle is$\;x^2+y^2-6x+2y+c=0$$\Rightarrow \;\; (x-3)^2+(y+1)^2=10-c$So, the center of the circle is $C(3,-1)$ and the radius is $r^2=10-C$Now, since the equation of tangent is$x-2y=0$Slop, $m_t=\frac{1}{2}$Slope of the radius, connecting the centre $(3,-1)$ to the point of tangency$P(x_p,y_p) \; \text{ is } \; m_r=\;\frac{y_p-(-1)}{x_p-3}=\;\frac{y_p+1}{x_p-3}$Now, since radius is perpendicular to the tangent, so$\; m_t \cdot m_r = -1$$\Rightarrow \;\frac{1}{2} \cdot \;\frac{y_p+1}{x_p-3} = -1 \Rightarrow y_p+1 = -2x_p+6$$\Rightarrow y_p = -2x_p+5$But, $P$ lies on the tangent line, its coordinates satisfy $x_p-2y_p=0$$\Rightarrow \;\; x_p = 2y_p$So, $\;\; y_p = -2x_p+5$$\Rightarrow \;\; -2(2y_p)+5 = -4y_p+5$$\Rightarrow \;\; 5y_p = 5 \Rightarrow y_p=1$So, $x_p = 2y_p = 2 \times 1 = 2$So, the point $P$ is $(2,1)$.Now, the distance between $(6,3)$ and $(2,1)$ is$\; d = \sqrt{(2-6)^2+(1-3)^2}$$\Rightarrow \sqrt{(-4)^2+(-2)^2} \Rightarrow \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}$$\therefore$ The distance of the point $(6,3)$ from $P$ is $2\sqrt{5}$.