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]
- A
Statement $1$ is false; Statement $2$ is true
- B
Statement $1$ is true; Statement $2$ is true, Statement $2$ is a correct explanation for Statement $1$.
- C
Statement $1$ is true; Statement $2$ is false
- D
Statement $1$ is true; Statement $2$ is true; Statement $2$ is not a correct explanation for Statement $1$.
Similar Questions
If the centre of a circle is $(-6, 8)$ and it passes through the origin, then equation to its tangent at the origin, is
If the centre of a circle is $(-6, 8)$ and it passes through the origin, then equation to its tangent at the origin, is
The normal to the circle ${x^2} + {y^2} - 3x - 6y - 10 = 0$at the point $(-3, 4)$, is
The normal to the circle ${x^2} + {y^2} - 3x - 6y - 10 = 0$at the point $(-3, 4)$, is
Let $A B$ be a chord of length $12$ of the circle $(x-2)^{2}+(y+1)^{2}=\frac{169}{4}$ If tangents drawn to the circle at points $A$ and $B$ intersect at the point $P$, then five times the distance of point $P$ from chord $AB$ is equal to$.......$
Let $A B$ be a chord of length $12$ of the circle $(x-2)^{2}+(y+1)^{2}=\frac{169}{4}$ If tangents drawn to the circle at points $A$ and $B$ intersect at the point $P$, then five times the distance of point $P$ from chord $AB$ is equal to$.......$
- [JEE MAIN 2022]
Number of integral points interior to the circle $x^2 + y^2 = 10$ from which exactly one real tangent can be drawn to the curve $\sqrt {{{\left( {x + 5\sqrt 2 } \right)}^2} + {y^2}} \, - \sqrt {{{\left( {x - 5\sqrt 2 } \right)}^2} + {y^2}\,} \, = 10$ are (where integral point $(x, y)$ means $x, y \in I)$
Number of integral points interior to the circle $x^2 + y^2 = 10$ from which exactly one real tangent can be drawn to the curve $\sqrt {{{\left( {x + 5\sqrt 2 } \right)}^2} + {y^2}} \, - \sqrt {{{\left( {x - 5\sqrt 2 } \right)}^2} + {y^2}\,} \, = 10$ are (where integral point $(x, y)$ means $x, y \in I)$
Tangent to the circle $x^2 + y^2$ = $5$ at the point $(1, -2)$ also touches the circle $x^2 + y^2 -8x + 6y + 20$ = $0$ . Then its point of contact is
Tangent to the circle $x^2 + y^2$ = $5$ at the point $(1, -2)$ also touches the circle $x^2 + y^2 -8x + 6y + 20$ = $0$ . Then its point of contact is