JEE MAIN - Mathematics (2024 - 31st January Morning Shift - No. 19)
If one of the diameters of the circle $$x^2+y^2-10 x+4 y+13=0$$ is a chord of another circle $$\mathrm{C}$$, whose center is the point of intersection of the lines $$2 x+3 y=12$$ and $$3 x-2 y=5$$, then the radius of the circle $$\mathrm{C}$$ is :
4
3$$\sqrt2$$
6
$$\sqrt{20}$$
Explanation
_31st_January_Morning_Shift_en_19_1.png)
$$\begin{aligned} & 2 x+3 y=12 \\ & 3 x-2 y=5 \end{aligned}$$
$$\begin{aligned} & 13 x=39 \\ & x=3, y=2 \end{aligned}$$
Center of given circle is $$(5,-2)$$
Radius $$\sqrt{25+4-13}=4$$
$$\begin{aligned} & \therefore \mathrm{CM}=\sqrt{4+16}=5 \sqrt{2} \\ & \therefore \mathrm{CP}=\sqrt{16+20}=6 \end{aligned}$$
Comments (0)
