Q.Find the equation of the circle whose centre lies on the line $3x + 4y = 5$ and passes through the points $(1, -2)$ and $(4, -3)$.

Showing the 12 most recent of 14 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.The distance from x-axis to point (x, y, z) will be:

   (a) \sqrt{x^2+y^2}

   (b) \sqrt{x^2+z^2}

   (c) \sqrt{y^2+z^2}

   (d) \sqrt{x^2+y^2+z^2}

   ›Reveal solutionSolution

   The distance from a point to the $x$-axis is the distance to its foot of perpendicular on the axis, which only involves the $y$ and $z$ coordinates.

   Concept: The $x$-axis consists of all points of the form $(t, 0, 0)$. The perpendicular distance from a point $P(x,y,z)$ to the $x$-axis is the distance from $P$ to its foot of perpendicular $M(x,0,0)$ on the axis (since $M$ has the same $x$-coordinate).

   Working: …

   🔒 Start your 14-day free trial to see the full solution →
2. CBSE 2024Set 65/1/11 markMCQ
   Q.The distance of point $P(a, b, c)$ from the $y$-axis is: (A) $b$ (B) $b^2$ (C) $\sqrt{a^2 + c^2}$ (D) $a^2 + c^2$

   ›Reveal solutionSolution

   The distance from a point to the $y$-axis is the length of the perpendicular from the point to the axis; since the $y$-axis is the set of all points $(0, y, 0)$, we measure how far the point is in the $xz$-plane, giving $\sqrt{a^2 + c^2}$.

   Understanding Distance from an Axis

   When we talk about the distance of a point from an axis in three-dimensional space, we mean the perpendicular distance—the shortest straight-line path from the point to that axis.

   The $y$-axis consists of all points of the form $(0, y, 0)$ where $y$ can be any real number. Think of it as a vertical line running through the origin, where both the $x$-coordinate and $z$-coordinate are always zero.

   For a point $P(a, b, c)$, finding the distance to the $y$-axis means finding the closest point on the $y$-axis and measuring that distance. The closest point on the $y$-axis to $P$ will have the same $y$-coordinate as $P$ (namely $b$), but will have $x = 0$ and $z = 0$. So the closest point is $Q(0, b, 0)$.

   Step-by-Step Solution

   1. Identify the closest point on the $y$-axis.

      The point on the $y$-axis nearest to $P(a, b, c)$ is $Q(0, b, 0)$. This is because the perpendicular from $P$ to the $y$-axis drops straight down in the $xz$-plane while maintaining the same $y$-coordinate.

   2. Apply the distance formula in 3D.

      The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is:

   $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

   3. Calculate the distance from $P$ to $Q$.

      Substituting $P(a, b, c)$ and $Q(0, b, 0)$:

   $d = \sqrt{(0 - a)^2 + (b - b)^2 + (0 - c)^2}$

   $d = \sqrt{a^2 + 0 + c^2}$

   $d = \sqrt{a^2 + c^2}$

   4. Interpret geometrically. …

   🔒 Start your 14-day free trial to see the full solution →
3. CBSE 2024Set 65/2/11 markMCQ
   Q.The coordinates of the foot of the perpendicular drawn from the point $(0, 1, 2)$ on the x-axis are given by: (A) $(1, 0, 0)$ (B) $(2, 0, 0)$ (C) $(\sqrt{5}, 0, 0)$ (D) $(0, 0, 0)$

   ›Reveal solutionSolution

   The foot of the perpendicular from a point to an axis is its projection onto that axis. For the x-axis, this means setting the y and z coordinates to zero while keeping the x-coordinate. Thus, for $(0, 1, 2)$, the foot of the perpendicular on the x-axis is $(0, 0, 0)$.

   When we talk about the "foot of the perpendicular" from a point to a line (or an axis in 3D space), we are essentially looking for the point on that line which is closest to the given point. Geometrically, if you imagine dropping a plumb line from the given point straight down to the axis, where it touches the axis is the foot of the perpendicular. This concept is also known as the orthogonal projection of the point onto the line.

   Consider a point $P(x_0, y_0, z_0)$ in 3D space.

   The x-axis is a line where all points have their y-coordinate and z-coordinate equal to zero. Any point on the x-axis can be represented as $(x, 0, 0)$.

   When we drop a perpendicular from $P(x_0, y_0, z_0)$ to the x-axis, the resulting point on the x-axis, let's call it $F$, will have its y and z coordinates equal to $0$. The key insight is that the x-coordinate of $F$ will be exactly the same as the x-coordinate of $P$. This is because moving along the y or z directions does not change the x-position relative to the x-axis.

   Think of it this way: to find the point on the x-axis closest to $P(x_0, y_0, z_0)$, we need to eliminate the "deviation" in the y and z directions. We do this by setting $y=0$ and $z=0$. The x-coordinate remains untouched because the x-axis itself extends along the x-direction.

   The foot of the perpendicular from a point $P(x_0, y_0, z_0)$ to the x-axis is $F(x_0, 0, 0)$.

   Similarly, for the y-axis, it is $F(0, y_0, 0)$, and for the z-axis, it is $F(0, 0, z_0)$.

   Let's apply this understanding to the given problem.

   1. Identify the given point: We are given the point $P(0, 1, 2)$. Here, $x_0 = 0$, $y_0 = 1$, and $z_0 = 2$.

   2. Identify the target axis: We need to find the foot of the perpendicular on the x-axis. …

   🔒 Start your 14-day free trial to see the full solution →
4. CBSE 2024Set ANNUAL1 markMCQ
   Q.The distance of the point $P(a,b,c)$ from the $x$-axis is

   (a) $\sqrt{b^2+c^2}$

   (b) $\sqrt{a^2+c^2}$

   (c) $\sqrt{a^2+b^2}$

   (d) None of these

   ›Reveal solutionSolution

   The foot of perpendicular from P(a,b,c) to the x-axis is (a,0,0); the distance formula then leaves only b and c.

   …

   🔒 Start your 14-day free trial to see the full solution →
5. CBSE 2020Set 65/1/11 markMCQ
   Q.The length of the perpendicular drawn from the point $(4, -7, 3)$ to the $y$-axis will be (A) 3 units (B) 4 units (C) 5 units (D) 7 units

   ›Reveal solutionSolution

   The distance from a point to the $y$-axis is the perpendicular distance in the $xz$-plane, ignoring the $y$-coordinate. For $(4, -7, 3)$, this distance is $\sqrt{4^2 + 3^2} = 5$ units. The correct option is (C).

   The key idea: the $y$-axis is the set of all points where $x = 0$ and $z = 0$, with $y$ free. So the perpendicular from any point to the $y$-axis lands at the point that keeps the same $y$-coordinate but sets $x$ and $z$ to zero. The distance is then just the straight-line distance in the $xz$-plane from $(x, z)$ to $(0, 0)$.

   Think of it this way: if you stand at $(4, -7, 3)$ and walk straight toward the $y$-axis, you move only in the $x$ and $z$ directions — your $y$ doesn't change because the axis runs parallel to your $y$ direction. The shortest path is the hypotenuse of a right triangle with legs $4$ (the $x$-distance) and $3$ (the $z$-distance).

   1. Identify the foot of the perpendicular.

      The $y$-axis consists of points $(0, y, 0)$. The perpendicular from $(4, -7, 3)$ to this axis will have the same $y$-coordinate as the point, because the axis is vertical in $y$. So the foot is $(0, -7, 0)$.

   2. Compute the distance between the point and the foot.

      Use the 3D distance formula:

   $\text{Distance} = \sqrt{(4 - 0)^2 + (-7 - (-7))^2 + (3 - 0)^2}$

   The $y$-difference is $0$, so it simplifies to:

   $\sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$

   3. Interpret geometrically. …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2020Set 65/2/11 markMCQ
   Q.The coordinates of the foot of the perpendicular drawn from the point $(2, -3, 4)$ on the y-axis is (A) $(2, 3, 4)$ (B) $(-2, -3, -4)$ (C) $(0, -3, 0)$ (D) $(2, 0, 4)$

   ›Reveal solutionSolution

   The foot of the perpendicular from any point to the y-axis has the form $(0, y, 0)$; we preserve only the y-coordinate of the original point. The answer is $(0, -3, 0)$.

   The y-axis is the set of all points of the form $(0, y, 0)$ where $y$ ranges over all real numbers. When we drop a perpendicular from a point in space to the y-axis, we're finding the closest point on that axis.

   The key insight is geometric: a perpendicular from a point to a coordinate axis lands at the point on that axis which shares the same coordinate as the original point along that axis direction. Think of it as projecting the point straight onto the axis, collapsing the other two dimensions.

   For the y-axis specifically, we keep the y-coordinate and set both x and z to zero.

   1. Identify the structure of points on the y-axis

      Any point on the y-axis has coordinates $(0, y, 0)$ for some value of $y$. The x and z coordinates are always zero.

   2. Understand what "foot of the perpendicular" means

      The foot of the perpendicular from point $P = (2, -3, 4)$ to the y-axis is the point $Q$ on the y-axis such that the line segment $PQ$ is perpendicular to the y-axis.

   3. Apply the perpendicularity condition …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2020Set 65/3/11 markMCQ
   Q.The coordinates of the foot of the perpendicular drawn from the point $(-2, 8, 7)$ on the XZ-plane is (A) $(-2, -8, 7)$ (B) $(2, 8, -7)$ (C) $(-2, 0, 7)$ (D) $(0, 8, 0)$

   ›Reveal solutionSolution

   The foot of the perpendicular from any point to the XZ-plane is found by setting the $y$-coordinate to zero while keeping $x$ and $z$ unchanged; the answer is $(-2, 0, 7)$.

   The XZ-plane is the set of all points where $y = 0$. Think of it as the "floor" that contains the $x$-axis and $z$-axis but has no height in the $y$-direction. When we drop a perpendicular from a point in space to this plane, we're asking: what point on the XZ-plane is closest to our given point?

   The key insight is that a perpendicular to the XZ-plane must be parallel to the $y$-axis (since the XZ-plane is perpendicular to the $y$-axis). So dropping a perpendicular means moving straight up or down in the $y$-direction until we hit the plane where $y = 0$. The $x$ and $z$ coordinates don't change during this vertical drop.

   Let's work through this systematically.

   1. Identify the given point and the target plane.

      We have the point $P = (-2, 8, 7)$ and we need to find where the perpendicular from $P$ meets the XZ-plane.

   2. Recall the equation of the XZ-plane.

      The XZ-plane consists of all points $(x, y, z)$ satisfying $y = 0$. This is a plane perpendicular to the $y$-axis.

   3. Understand the geometry of the perpendicular.

      A line perpendicular to the XZ-plane must be parallel to the normal vector of that plane. The normal to the XZ-plane is $\hat{j} = (0, 1, 0)$, pointing in the $y$-direction. Therefore, the perpendicular from $P$ is a vertical line (in the $y$-sense) passing through $P$. …

   🔒 Start your 14-day free trial to see the full solution →
8. CBSE 2019Set ANNUAL1 markMCQ
   Q.Shortest distance between Point (a, b, c) and y-axis is:

   (a) √(a²+b²)

   (b) √(b²+c²)

   (c) √(c²+a²)

   (d) √(a²+b²+c²)

   ›Reveal solutionSolution

   Every point on the $y$-axis has the form $(0,t,0)$; the perpendicular distance from $(a,b,c)$ to the axis ignores the $y$-coordinate.

   The foot of the perpendicular from $(a,b,c)$ to the $y$-axis is $(0,b,0)$, so the shortest distance is …

   🔒 Start your 14-day free trial to see the full solution →
9. CBSE 2019Set ANNUAL1 markMCQ
   Q.The distance of the point $(3, 4, 5)$ from $x$-axis is -

   (a) $3$

   (b) $5$

   (c) $\sqrt{41}$

   (d) None of these

   ›Reveal solutionSolution

   Distance $=\sqrt{y^2+z^2}=\sqrt{41}$.

   …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2018Set ANNUAL1 markMCQ
    Q.The distance of point $(3, 4, 5)$ from XZ-plane is

    (a) $4$

    (b) $3$

    (c) $5$

    (d) $0$

    ›Reveal solutionSolution

    Distance from the XZ-plane is $|y|=4$.

    …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2018Set ANNUAL1 markMCQ
    Q.The shortest distance of point $(a, b, c)$ from X-axis is:

    (a) $\sqrt{b^2+c^2}$

    (b) $\sqrt{a^2+b^2}$

    (c) $\sqrt{c^2+a^2}$

    (d) $\sqrt{a^2+b^2+c^2}$

    ›Reveal solutionSolution

    Distance of $(a,b,c)$ from the X-axis is $\sqrt{b^2+c^2}$.

    A general point on the X-axis is $(t,0,0)$. The nearest such point to $(a,b,c)$ is $(a,0,0)$, so the shortest distance …

    🔒 Start your 14-day free trial to see the full solution →
12. CBSE 2018Set ANNUAL1 mark
    Q.Express the Ellipse $(a-b)x^2 + (a+b)y^2 = a^2 - b^2$ in standard form.

    ›Reveal solutionSolution

    Divide the whole equation by a²−b² = (a−b)(a+b) to get x²/(a+b) + y²/(a−b) = 1.

    We start with (a−b)x² + (a+b)y² = a²−b².

    Step 1: Factor the right side. a²−b² = (a−b)(a+b).

    Step 2: Divide both sides by (a−b)(a+b):

    (a−b)x² / [(a−b)(a+b)] + (a+b)y² / [(a−b)(a+b)] = 1

    Step 3: Cancel the common factors: …

    🔒 Start your 14-day free trial to see the full solution →