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^{2}+ y^{2}+2x + 2ky + 6 = 0 and x^{2}+ y^{2}+ 2ky + k = 0 intersect orthogonally then k equals ?^{2}+ y^{2}+ 2gx + 2fy + c = 0. Then the equation of the circumcircle of the triangle OPQ is^{2}+ y^{2}= 2 is ?^{2}+ y^{2}= r^{2}at A and B. Then PA.PB is equal to ?^{2}+ y^{2}= 9 is^{2}+ y^{2}+ x + = 0 and x^{2}+ y^{2}- x = 0 are^{2}+ y^{2}- 6x + 2y - 54 = 0 is ?^{2}+ y^{2}+ 4x + 2y - 4 = 0 is reflected in a mirror to become the circle x^{2}+ y^{2}+ 6x + 4y + 4 = 0, the mirror is lying along the straight line^{2}+ y^{2}+ 4x + 22y + m = 0 bisects the circumference of the circle x^{2}+ y^{2}- 2x + 8y - n = 0 then m + n = ?^{2}+ y^{2}- 12x - 4y + 30 = 0 which is farthest from the origin are$\overrightarrow{r}=(8+3\lambda )\hat{i}-(9+16\lambda )\hat{j}+(10+7\lambda )\hat{k}and\overrightarrow{r}=15\hat{i}+29\hat{j}+5\hat{k}+\mu (3\hat{i}+8\hat{j}-5\hat{k})$

x + y + z= 7 is(A) $\sqrt{\frac{2}{3}}$ (B) $\frac{2}{\sqrt{3}}$ (C) $\frac{2}{3}$ (D) $\frac{1}{3}$

^{2}+ y^{2}- 4x - 6y + 10 = 0 orthogonally, the equation of the circle isii) $\overrightarrow{r}=\left(3-t\right)\hat{i}+\left(4+2t\right)\hat{j}+\left(t-2\right)\hat{k}$

$\overrightarrow{r}=\left(1+s\right)\hat{i}+\left(3s-7\right)\hat{j}+\left(2s-2\right)\hat{k}$

A) x + y = 0 B) x + 3y = 0 C) x = y D) 3x + 2y = 0

^{2}+ y^{2}= a^{2}may subtend a right angle at the centre of the circle is^{2}+ y^{2}+ 2a_{i}x = c (i = 1,2,3) are in G.P, then the lengths of the tangents drawn to them from any point on the circle x^{2}+ y^{2}= c^{2}are in^{2}+ y^{2}+ x + y = 0 and x^{2}+ y^{2}+ x - y = 0 intersect at an angle?