Tangents drawn from origin to the circle ${x^2} + {y^2} - 2ax - 2by + {b^2} = 0$ are perpendicular to each other, if
Tangents drawn from origin to the circle ${x^2} + {y^2} - 2ax - 2by + {b^2} = 0$ are perpendicular to each other, if
- A
$a - b = 1$
- B
$a + b = 1$
- C
${a^2} = {b^2}$
- D
${a^2} + {b^2} = 1$
Similar Questions
The equations of the tangents drawn from the origin to the circle ${x^2} + {y^2} - 2rx - 2hy + {h^2} = 0$ are
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- [IIT 1988]
Statement $1$ : The only circle having radius $\sqrt {10} $ and a diameter along line $2x + y = 5$ is $x^2 + y^2 - 6x +2y = 0$.
Statement $2$ : $2x + y = 5$ is a normal to the circle $x^2 + y^2 -6x+2y = 0$.
Statement $1$ : The only circle having radius $\sqrt {10} $ and a diameter along line $2x + y = 5$ is $x^2 + y^2 - 6x +2y = 0$.
Statement $2$ : $2x + y = 5$ is a normal to the circle $x^2 + y^2 -6x+2y = 0$.
- [JEE MAIN 2013]
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The equation of the tangent to the circle ${x^2} + {y^2} = {a^2}$ which makes a triangle of area ${a^2}$ with the co-ordinate axes, is
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The tangent$(s)$ from the point of intersection of the lines $2x -3y + 1$ = $0$ and $3x -2y -1$ = $0$ to circle $x^2 + y^2 + 2x -4y$ = $0$ will be -
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