JEE Main Circle Practice Questions With Solutions

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

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

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

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

A circle is inscribed in an equilateral triangle of side of length 12. If the area and perimeter of any square inscribed in this circle are $m$m and $n$n, respectively, then $m+n^2$m + n^{2} is equal to

Let the circle $C_1:x^2+y^2-2(x+y)+1=0$C_{1} : x^{2} + y^{2} - 2 \left(\right. x + y \left.\right) + 1 = 0 and ${\mathrm{C}}_2$C_{2} be a circle having centre at $(-1,0)$\left(\right. - 1 , 0 \left.\right) and radius 2 . If the line of the common chord of ${\mathrm{C}}_1$C_{1} and ${\mathrm{C}}_2$C_{2} intersects the $\mathrm{y}$y-axis at the point $\mathrm{P}$P, then the square of the distance of P from the centre of ${\mathrm{C}}_1$C_{1} is:

Let ABCD and AEFG be squares of side 4 and 2 units, respectively. The point E is on the line segment AB and the point F is on the diagonal AC. Then the radius r of the circle passing through the point F and touching the line segments BC and CD satisfies :

Let a circle C of radius 1 and closer to the origin be such that the lines passing through the point $(3,2)$\left(\right. 3 , 2 \left.\right) and parallel to the coordinate axes touch it. Then the shortest distance of the circle C from the point $(5,5)$\left(\right. 5 , 5 \left.\right) is :

Let $\mathrm{C}$C be a circle with radius $\sqrt{10}$\sqrt{10} units and centre at the origin. Let the line $x+y=2$x + y = 2 intersects the circle $\mathrm{C}$C at the points $\mathrm{P}$P and $\mathrm{Q}$Q. Let $\mathrm{MN}$MN be a chord of $\mathrm{C}$C of length 2 unit and slope $-1$- 1. Then, a distance (in units) between the chord PQ and the chord $\mathrm{MN}$MN is

A square is inscribed in the circle $x^2+y^2-10x-6y+30=0$x^{2} + y^{2} - 10 x - 6 y + 30 = 0. One side of this square is parallel to $y=x+3$y = x + 3. If $(x_i,y_i)$\left(\right. x_{i} , y_{i} \left.\right) are the vertices of the square, then $\mathrm{\Sigma }(x_i^2+y_i^2)$\Sigma \left(\right. x_{i}^{2} + y_{i}^{2} \left.\right) is equal to: