Question

1. Find the equation of the normal to the circle $x^2 + y^2 = 2x$, which is parallel to the line $x + 2y = 3$.
2. Find the equation of normal to the circle $x^2 + y^2 - 2x - 2y + 1 = 0$ and passing through (-1,2).
3. Find the equation of normal to the circle $2x^2 + 2y^2 - 4x + 6y - 2 = 0$ which is at maximum distance from (2,3).
4. Find the equation of common normal of the following circles (i) $x^2 + y^2 = 4$ and $x^2 + y^2 - 4x + 2y - 1 = 0$ (ii) $(x - 2)(x - 4) + (y - 1)(y - 5) = 0$ and $4x - 6y - x^2 - y^2 = 1$ (iii) $x = 1 + 2cos\theta, y = 2 + 2 sin\theta$, where $\theta$ is parameter $(x - 1)^2 + y^2 = 4y + 2$.
5. Statement I : The only circle having radius $\sqrt{10}$ and a diameter along line $2x + y = 5$ is $x^2 + y^2 - 6x + 2y = 0$. Statement II : $2x + y = 5$ is a normal to the circle $x^2 + y^2 - 6x + 2y = 0$.

• A. Statement I is false, Statement II is true
• B. Statement I is true; Statement II is false.
• C. Statement I is true, Statement II is true, Statement II is not a correct explanation for Statement I.
• D. Statement I is true : Statement II is true; Statement II is a correct explanation for Statement I.

Answer 1

Answer: (A)

(1) Statement I is false, Statement II is true

Answer 2

Problem 1: The circle is $x^2 + y^2 - 2x = 0$, which is $(x-1)^2 + y^2 = 1$. The center is $(1,0)$. Normals to a circle pass through its center. The line $x+2y=3$ has a slope of $-\frac{1}{2}$. Since the normal is parallel to this line, its slope is also $-\frac{1}{2}$. The equation of the normal is $y-0 = -\frac{1}{2}(x-1)$, which simplifies to $x+2y=1$.

Problem 2: The circle is $x^2 + y^2 - 2x - 2y + 1 = 0$, which is $(x-1)^2 + (y-1)^2 = 1$. The center is $(1,1)$. A normal passes through the center. The normal also passes through $(-1,2)$. The slope of the line joining $(1,1)$ and $(-1,2)$ is $\frac{2-1}{-1-1} = \frac{1}{-2} = -\frac{1}{2}$. The equation of the normal is $y-1 = -\frac{1}{2}(x-1)$, which simplifies to $x+2y=3$.

Problem 3: The circle is $2x^2 + 2y^2 - 4x + 6y - 2 = 0$, which simplifies to $x^2 + y^2 - 2x + 3y - 1 = 0$. The center is $(1, -\frac{3}{2})$. Normals pass through the center. The distance from $(2,3)$ to a normal is maximized when the normal is perpendicular to the line joining the center $(1, -\frac{3}{2})$ and the point $(2,3)$. The slope of the line joining these points is $\frac{3 - (-\frac{3}{2})}{2 - 1} = \frac{9/2}{1} = \frac{9}{2}$. The slope of the normal is the negative reciprocal, $-\frac{2}{9}$. The equation of the normal is $y - (-\frac{3}{2}) = -\frac{2}{9}(x - 1)$, which simplifies to $4x + 18y + 23 = 0$.

Problem 4 (i): Circle 1: $x^2 + y^2 = 4$, center $(0,0)$. Circle 2: $x^2 + y^2 - 4x + 2y - 1 = 0$, center $(2,-1)$. A common normal must pass through both centers. Since the centers are distinct, there is no common normal.

Problem 4 (ii): Circle 1: $(x - 2)(x - 4) + (y - 1)(y - 5) = 0 \implies x^2 + y^2 - 6x - 6y + 13 = 0$. Center $(3,3)$. Circle 2: $4x - 6y - x^2 - y^2 = 1 \implies x^2 + y^2 - 4x + 6y + 1 = 0$. Center $(2,-3)$. The common normal is the line passing through both centers $(3,3)$ and $(2,-3)$. The slope is $\frac{-3-3}{2-3} = 6$. The equation is $y-3 = 6(x-3)$, which simplifies to $6x - y - 15 = 0$.

Problem 4 (iii): Circle 1: $x = 1 + 2\cos\theta, y = 2 + 2 \sin\theta \implies (x-1)^2 + (y-2)^2 = 4$. Center $(1,2)$. Circle 2: $(x - 1)^2 + y^2 = 4y + 2 \implies (x-1)^2 + (y-2)^2 = 6$. Center $(1,2)$. Both circles are concentric with center $(1,2)$. Any line passing through the common center is a common normal. The equation $x=1$ is a vertical line passing through $(1,2)$.

Problem 5: Statement I: The circle $x^2 + y^2 - 6x + 2y = 0$ has center $(3,-1)$ and radius $\sqrt{10}$. The line $2x+y=5$ passes through $(3,-1)$, so it is a diameter. However, there are infinitely many circles with radius $\sqrt{10}$ and a diameter along $2x+y=5$, as any point on $2x+y=5$ can be the center. Thus, Statement I is false. Statement II: The line $2x+y=5$ passes through the center $(3,-1)$ of the circle $x^2 + y^2 - 6x + 2y = 0$. Therefore, it is a normal to the circle. Statement II is true. Since Statement I is false and Statement II is true, option (1) is correct.