Tangents AB and AC are drawn from the point A | vedclass

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

Question

(b) Equation of $BC$ (chord of contact) is

$0.x + 1.y - (x + 0) + 2(y + 1) + 1 = 0$ or $ - x + 3y + 3 = 0$

Equation of circle through $B$ and $C

Vedclass · Accepted Answer

${x^2} + {y^2} - x + y - 2 = 0$

Vedclass · Answer

(b) Equation of $BC$ (chord of contact) is

$0.x + 1.y - (x + 0) + 2(y + 1) + 1 = 0$ or $ - x + 3y + 3 = 0$

Equation of circle through $B$ and $C$ i.e., intersection of the given circle and chord of contact is

$({x^2} + {y^2} - 2x + 4y + 1) + \lambda ( - x + 3y + 3) = 0$.

It passes through $A(0,\;1)$, so the equation of the required circle is ${x^2} + {y^2} - x + y - 2 = 0$.

Aliter : Centre of the required circle is mid-point of $A(0,\;1)$ and centre of the given circle i.e., $(1,\; - 2)$.

Therefore, centre $\left( {\frac{1}{2},\; - \frac{1}{2}} \right)$ and radius $\sqrt {\frac{5}{2}} $.

Hence the circle is ${x^2} + {y^2} - x + y - 2 = 0$.

Tangents $AB$ and $AC$ are drawn from the point $A(0,\,1)$ to the circle ${x^2} + {y^2} - 2x + 4y + 1 = 0$. Equation of the circle through $A, B$ and $C$ is

Similar Questions

The equations of the tangents to the circle ${x^2} + {y^2} - 6x + 4y = 12$ which are parallel to the straight line $4x + 3y + 5 = 0$, are

The equation of the normal to the circle ${x^2} + {y^2} - 2x = 0$ parallel to the line $x + 2y = 3$ is

Square of the length of the tangent drawn from the point $(\alpha ,\beta )$ to the circle $a{x^2} + a{y^2} = {r^2}$ is

If the line $lx + my + n = 0$ be a tangent to the circle ${(x - h)^2} + {(y - k)^2} = {a^2},$ then

The equations of the tangents drawn from the point $(0, 1)$ to the circle ${x^2} + {y^2} - 2x + 4y = 0$ are