1
JEE Main 2024 (Online) 9th April Morning Shift
+4
-1

Let a circle passing through $$(2,0)$$ have its centre at the point $$(\mathrm{h}, \mathrm{k})$$. Let $$(x_{\mathrm{c}}, y_{\mathrm{c}})$$ be the point of intersection of the lines $$3 x+5 y=1$$ and $$(2+\mathrm{c}) x+5 \mathrm{c}^2 y=1$$. If $$\mathrm{h}=\lim _\limits{\mathrm{c} \rightarrow 1} x_{\mathrm{c}}$$ and $$\mathrm{k}=\lim _\limits{\mathrm{c} \rightarrow 1} y_{\mathrm{c}}$$, then the equation of the circle is :

A
$$5 x^2+5 y^2-4 x-2 y-12=0$$
B
$$25 x^2+25 y^2-20 x+2 y-60=0$$
C
$$25 x^2+25 y^2-2 x+2 y-60=0$$
D
$$5 x^2+5 y^2-4 x+2 y-12=0$$
2
JEE Main 2024 (Online) 8th April Evening Shift
+4
-1

If the image of the point $$(-4,5)$$ in the line $$x+2 y=2$$ lies on the circle $$(x+4)^2+(y-3)^2=r^2$$, then $$r$$ is equal to:

A
2
B
3
C
4
D
1
3
JEE Main 2024 (Online) 8th April Morning Shift
+4
-1

Let the circles $$C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2$$ and $$C_2:(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2$$ touch each other externally at the point $$(6,6)$$. If the point $$(6,6)$$ divides the line segment joining the centres of the circles $$C_1$$ and $$C_2$$ internally in the ratio $$2: 1$$, then $$(\alpha+\beta)+4\left(r_1^2+r_2^2\right)$$ equals

A
130
B
110
C
145
D
125
4
JEE Main 2024 (Online) 6th April Evening Shift
+4
-1

If $$\mathrm{P}(6,1)$$ be the orthocentre of the triangle whose vertices are $$\mathrm{A}(5,-2), \mathrm{B}(8,3)$$ and $$\mathrm{C}(\mathrm{h}, \mathrm{k})$$, then the point $$\mathrm{C}$$ lies on the circle :

A
$$x^2+y^2-74=0$$
B
$$x^2+y^2-65=0$$
C
$$x^2+y^2-61=0$$
D
$$x^2+y^2-52=0$$
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