The number of common tangents that can be dra | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given equations of the circles are $x^2+y^2-2x-2y-23=0 \dots (1)$ and $x^2+y^2-4x-4y-1=0 \dots (2)$.For circle $(1)$,the centre $C_1 = (1, 1)$

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) The given equations of the circles are $x^2+y^2-2x-2y-23=0 \dots (1)$ and $x^2+y^2-4x-4y-1=0 \dots (2)$.For circle $(1)$,the centre $C_1 = (1, 1)$ and radius $r_1 = \sqrt{1^2+1^2-(-23)} = \sqrt{25} = 5$.For circle $(2)$,the centre $C_2 = (2, 2)$ and radius $r_2 = \sqrt{2^2+2^2-(-1)} = \sqrt{9} = 3$.The distance between the centres $C_1$ and $C_2$ is $d = \sqrt{(2-1)^2+(2-1)^2} = \sqrt{1+1} = \sqrt{2} \approx 1.414$.We compare $d$ with the difference of the radii $|r_1 - r_2| = |5 - 3| = 2$.Since $d Therefore,the number of common tangents that can be drawn is $0$.

The number of common tangents that can be drawn to the circles $x^2+y^2-2x-2y-23=0$ and $x^2+y^2-4x-4y-1=0$ is

Similar Questions

$A$ tangent $PT$ is drawn to the circle $x^2 + y^2 = 4$ at the point $P(\sqrt{3}, 1)$. $A$ line $L$ perpendicular to $PT$ is a tangent to the circle $(x - 3)^2 + y^2 = 1$. The possible equation of $L$ is:

The radical centre of the three circles $x^2+y^2-1=0$,$x^2+y^2-8x+15=0$ and $x^2+y^2+10y+24=0$ is

If $x-y+1=0$ meets the circle $x^2+y^2+y-1=0$ at $A$ and $B$,then the equation of the circle with $AB$ as diameter is

Find the equation of the radical axis of the two circles $x^2 + y^2 - x + 1 = 0$ and $3(x^2 + y^2) + y - 1 = 0$.

The equation of the circle having its centre on the line $x + 2y - 3 = 0$ and passing through the points of intersection of the circles ${x^2} + {y^2} - 2x - 4y + 1 = 0$ and ${x^2} + {y^2} - 4x - 2y + 4 = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module