The shortest distance from the point $(2, -7)$ to the circle $x^2 + y^2 - 14x - 10y - 151 = 0$ is equal to 5.

The shortest distance from the point (2, -7) to th – 1st PUC Mathematics

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(2, -7)

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Concept understanding — Circle Equation Standard Form

Where the Circle Equation Comes From

Imagine you're standing at a point on a flat field. You tie a rope to a stake at that point, walk out the full length of the rope, and start walking in a circle, keeping the rope taut. Every point you step on is exactly the same distance from the stake.

That's the entire idea: a circle is the set of all points that are a fixed distance (the radius) from a fixed point (the centre).

If we put this on a coordinate plane, we can turn that geometric idea into an algebraic equation.

From Geometry to Algebra

Let the centre be at coordinates $(h, k)$ . Let the radius be $r$ . Take any point $(x, y)$ that lies on the circle. The distance from $(x, y)$ to $(h, k)$ must equal $r$ .

What's the distance between two points in the plane? The distance formula:

\sqrt{(x - h)^2 + (y - k)^2} = r

Now square both sides to remove the square root:

(x - h)^2 + (y - k)^2 = r^2

That's it. That's the standard form of the equation of a circle.

(x - h)^2 + (y - k)^2 = r^2

What Each Piece Tells You

$(x - h)$ and $(y - k)$ — these shift the circle away from the origin. If the centre is at $(0,0)$ , the equation simplifies to $x^2 + y^2 = r^2$ .
$r^2$ — notice it's the square of the radius, not the radius itself. If the equation says $x^2 + y^2 = 25$ , the radius is $\sqrt{25} = 5$ , not 25.
The equals sign — only the points $(x, y)$ that make this equation true lie on the circle. Any other point gives a larger or smaller left-hand side.

Watch out

A common mistake: for $(x - 3)^2 + (y + 2)^2 = 16$ , students often read the centre straight off the signs printed in the equation and say $(3, 2)$ . That's wrong. Each bracket must first be written in the exact form $x - h$ and $y - k$ : here $(y + 2) = (y - (-2))$ , so $k = -2$ , not $2$ . The centre is actually $(3, -2)$ . Always flip the sign inside every bracket before reading off $h$ and $k$ .

Quick Example

Write the equation of a circle with centre $(-1, 4)$ and radius $3$ .

Here $h = -1$ , $k = 4$ , $r = 3$ . Plug in:

(x - (-1))^2 + (y - 4)^2 = 3^2

Simplify:

$(x + 1)^2 + (y - 4)^2 = 9$ …

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The center is $(7,5)$ and radius $15$ ; the point $(2,-7)$ lies $13$ units from the center, so it is inside the circle and the shortest distance is $15-13=2$ , not $5$ . The statement is false.

For a point and a circle, the shortest distance is measured along the line through the center.

1. Center and radius. Complete the square on $x^2+y^2-14x-10y-151=0$ :

(x-7)^2+(y-5)^2=151+49+25=225

So center $C=(7,5)$ and radius $r=\sqrt{225}=15$ .

2. Distance from the point to the center.

$d=\sqrt{(7-2)^2+(5+7)^2}=\sqrt{25+144}=\sqrt{169}=13$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2025Set 2025-M1 markMCQ
$2 x-3 y+12=0$
›Reveal solutionSolution
The circle's center is the intersection of its two given diameters, and its radius comes from the given area. This gives $x^2+y^2+6x-4y-36=0$ , matching option (D).

Step-by-step reasoning

Find the centre (intersection of the diameters).

$2x-3y+12=0, \qquad x+4y-5=0$
From the second: $x=5-4y$ . Substituting into the first:
$2(5-4y)-3y+12=0 \Rightarrow 10-8y-3y+12=0 \Rightarrow 22-11y=0 \Rightarrow y=2$
Then $x=5-4(2)=-3$ . Centre $=(-3,2)$ .

Find the radius from the area.

$\pi r^2 = 154 \quad (\pi\approx\tfrac{22}{7}) \quad\Rightarrow\quad r^2 = 154\times\frac{7}{22}=49 \quad\Rightarrow\quad r=7$

Write and expand the circle equation. $(x+3)^2+(y-2)^2=49$ …
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KCET 2021Set A-11 markMCQ
$y = \alpha x^2 - 6x + \beta$
›Reveal solutionSolution
The parabola passes through $(0,2)$ and has a horizontal tangent at $x = \frac{3}{2}$ . Using the point condition gives $\beta = 2$ , and the derivative condition gives $\alpha = 2$ . The correct pair is $\alpha = 2, \beta = 2$ , option (C).

The key idea here is that a tangent parallel to the $x$ -axis means the slope of the tangent is zero. For a curve $y = f(x)$ , the slope of the tangent at any point is given by $\frac{dy}{dx}$ . So "tangent parallel to $x$ -axis" translates directly to $\frac{dy}{dx} = 0$ at that $x$ -coordinate.

We also have a point that lies on the parabola. Substituting that point into the equation gives a direct relation between $\alpha$ and $\beta$ .

Let’s work through it step by step.

Use the point $(0,2)$ . The parabola is $y = \alpha x^2 - 6x + \beta$ . Substituting $x = 0$ , $y = 2$ :

$2 = \alpha(0)^2 - 6(0) + \beta \quad\Rightarrow\quad \beta = 2.$

Find the derivative. Differentiate $y$ with respect to $x$ :

$\frac{dy}{dx} = 2\alpha x - 6.$

Apply the tangent condition. At $x = \frac{3}{2}$ , the tangent is parallel to the $x$ -axis, so $\frac{dy}{dx} = 0$ :

$2\alpha \left(\frac{3}{2}\right) - 6 = 0 \quad\Rightarrow\quad 3\alpha - 6 = 0 \quad\Rightarrow\quad \alpha = 2.$

Check the options. We have $\alpha = 2$ , $\beta = 2$ . This matches option (C). …
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COMEDK 2021Set 20211 markMCQ
${x^2} + {y^2} - 2x - 4y + 1 = 0$
›Reveal solutionSolution

Check: centre = (1, 2), r = sqrt(1 + 4 - 1) = sqrt(4) = 2. Distance from centre to the X-axis = 2 = r, so it touches the X-axis. Correct.

Concept: A circle that touches the X-axis has radius equal to the absolute value of the y-coordinate of its centre (the perpendicular distance from the centre to the line y = 0).

Centre (h, k) = (1, 2) -> radius r = |k| = 2.

Equation: (x - 1)^2 + (y - 2)^2 = 2^2

x^2 - 2x + 1 + y^2 - 4y + 4 = 4

x^2 + y^2 - 2x - 4y + 1 = 0 …
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COMEDK 2021Set 20211 markMCQ
$2{x^2} + 2{y^2} + 3x + 4y + {9 \over 8} = 0$
›Reveal solutionSolution

The four options are single numbers, so they refer to the radius; the radius is 1.

Concept: general equation of a circle x^2 + y^2 + 2gx + 2fy + c = 0 has centre (-g, -f) and radius sqrt(g^2 + f^2 - c).

First make the coefficients of x^2 and y^2 equal to 1 by dividing throughout by 2:

2x^2 + 2y^2 + 3x + 4y + 9/8 = 0

=> x^2 + y^2 + (3/2)x + 2y + 9/16 = 0.

So 2g = 3/2 => g = 3/4; 2f = 2 => f = 1; c = 9/16.

Centre = (-3/4, -1). …
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COMEDK 2021Set 20211 markMCQ
${x^2} + {y^2} - 2x - 4y - 20 = 0$
›Reveal solutionSolution

Equation: (x - 1)^2 + (y - 2)^2 = 25 => x^2 - 2x + 1 + y^2 - 4y + 4 = 25 => x^2 + y^2 - 2x - 4y + 5 - 25 = 0 => x^2 + y^2 - 2x - 4y - 20 = 0.

Concept: circle with centre (h, k) and radius r: (x - h)^2 + (y - k)^2 = r^2, where r is the distance from the centre to any point on the circle.

Centre (1, 2), point on circle (4, 6).

r = sqrt((4 - 1)^2 + (6 - 2)^2) = sqrt(9 + 16) = sqrt(25) = 5.

Equation: (x - 1)^2 + (y - 2)^2 = 25 …
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KCET 2018Set A-11 markMCQ
$3x - 4y + 2 = 0$
›Reveal solutionSolution

Parallel lines have equal slopes, so keep the $x$ - and $y$ -coefficients unchanged and fix the constant by making the line pass through the given point.

Step 1 — Use the parallelism condition.

The slope of $3x - 4y + 2 = 0$ is
$m = -\frac{\text{coefficient of }x}{\text{coefficient of }y} = -\frac{3}{-4} = \frac{3}{4}$
Any line parallel to it has the same slope, so it can be written by changing only the constant term:
$3x - 4y + c = 0$
This is the standard "family of parallel lines" trick — it saves converting to slope-intercept form.

Step 2 — Impose the point condition.

The line must pass through $(-2, 3)$ , so those coordinates must satisfy it:
$3(-2) - 4(3) + c = 0$ $-6 - 12 + c = 0 \;\Rightarrow\; c = 18$
Step 3 — Write the equation.
$3x - 4y + 18 = 0$
Step 4 — Verify. …
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←Previous questionQ52 — The line x + 3y = 0 is a diameter of the circle x² + y² + 6x + 2y = 0.→Next questionQ54 — If the line lx + my = 1 is a tangent to the circle x² + y² = a², then …

Q.The shortest distance from the point (2,−7)(2, -7)(2,−7) to the circle x2+y2−14x−10y−151=0x^2 + y^2 - 14x - 10y - 151 = 0x2+y2−14x−10y−151=0 is equal to 5.

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Q.The shortest distance from the point $(2, -7)$ to the circle $x^2 + y^2 - 14x - 10y - 151 = 0$ is equal to 5.

Unlock everything free for 14 days