JEE Advanced 1988

LEVELJEE Main

Animated Solution for Mathematics - Circles: The equations of the tangents drawn from the origin to the circle $x^{2} + y^{2} - 2 r x - 2 h y + h^{2} = 0$ , are

Q: Can I use the discriminant method ($D=0$) instead of the distance formula?

Absolutely! You can substitute $y=mx$ into the circle equation to get a quadratic in $x$, then set the discriminant to zero. It is mathematically equivalent. However, the distance formula is often faster and less prone to messy algebraic errors. Choose the tool that feels most intuitive to you!

Select Answer:

* Multiple Correct

Visualized Solution

Analyze the Circle Equation

Given equation: $x^{2} + y^{2} - 2 r x - 2 h y + h^{2} = 0$
Let's group the terms to find the standard form: $(x^{2} - 2 r x) + (y^{2} - 2 h y + h^{2}) = 0$
Complete the square for $x$ by adding $r^{2}$ to both sides.
Standard Form: $(x - r)^{2} + (y - h)^{2} = r^{2}$

Identify Center and Radius

Comparing with standard circle equation: $(x - x_{0})^{2} + (y - y_{0})^{2} = R^{2}$
Center of the circle: $C = (r, h)$
Radius of the circle: $R = r$
Notice that the radius is exactly equal to the x-coordinate of the center.

Geometric Observation: The Y-Axis

The distance from the center $C (r, h)$ to the y-axis ( $x = 0$ ) is exactly $∣ r ∣ = r$ .
Since this perpendicular distance equals the radius of the circle, the y-axis must be tangent to the circle.
Therefore, our first tangent equation is: $x = 0$

Equation of the Second Tangent

Any line passing through the origin $O (0, 0)$ can be written as $y = m x$ .
Let's rewrite this in general form: $m x - y = 0$ .
We need to find the value of $m$ such that this line is tangent to our circle.

Condition for Tangency

The perpendicular distance from the center $C (r, h)$ to the line $m x - y = 0$ must equal the radius $r$ .
Using the distance formula: $d = \frac{∣ m ( r ) - ( h ) ∣}{m ^{2} + ( - 1 ) ^{2}}$
Setting this equal to the radius: $\frac{∣ m r - h ∣}{m ^{2} + 1} = r$

Squaring and Expanding

To solve for $m$ , let's square both sides to eliminate the absolute value and square root:
$\frac{( m r - h ) ^{2}}{m ^{2} + 1} = r^{2}$
Multiply both sides by $(m^{2} + 1)$ : $(m r - h)^{2} = r^{2} (m^{2} + 1)$
Expand both sides: $m^{2} r^{2} - 2 m r h + h^{2} = m^{2} r^{2} + r^{2}$

Isolating the Slope $m$

Notice that the term $m^{2} r^{2}$ appears on both sides and cancels out.
This leaves us with: $- 2 m r h + h^{2} = r^{2}$
Rearranging the terms to solve for $m$ : $2 m r h = h^{2} - r^{2}$
Thus, the slope is: $m = \frac{h ^{2} - r ^{2}}{2 r h}$

Final Equations of Tangents

Substitute the value of $m$ back into our line equation $y = m x$ :
$y = (\frac{h ^{2} - r ^{2}}{2 r h}) x$
Cross-multiplying gives: $2 r h y = (h^{2} - r^{2}) x$
Rearranging: $(h^{2} - r^{2}) x - 2 r h y = 0$
The two tangents are $x = 0$ and $(h^{2} - r^{2}) x - 2 r h y = 0$ .

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

The given equation is

x^{2} + y^{2} - 2 r x - 2 h y + h^{2} = 0

. To reveal the underlying geometry, we group the terms as

(x^{2} - 2 r x) + (y^{2} - 2 h y + h^{2}) = 0

By completing the square for the

x

terms, we add

r^{2}

to both sides. This transforms the equation into:

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

The chaos vanishes, revealing a circle with its center at

C (r, h)

and a radius of

r

. This is the foundation upon which we build our solution.

The Elegant Shortcut

Observe the relationship between the center

C (r, h)

and the radius

r

. The horizontal distance from the center to the

y

-axis is exactly

r

In geometry, when the perpendicular distance from the center to a line equals the radius, that line is a tangent. Without complex algebra, we identify the first tangent: the

y

-axis, defined by

x = 0

The Algebraic Hunt

We seek a second tangent passing through the origin

(0, 0)

. We represent this line as

y = m x

, or

m x - y = 0

To find the slope

m

, we apply the condition for tangency: the perpendicular distance from the center

C (r, h)

to the line

m x - y = 0

must equal the radius

r

. Using the distance formula, we obtain:

\frac{∣ m r - h ∣}{m ^{2} + 1} = r

The Final Calculation

To solve for

m

, we square both sides to eliminate the absolute value and the square root:

\frac{( m r - h ) ^{2}}{m ^{2} + 1} = r^{2}

Cross-multiplying yields

(m r - h)^{2} = r^{2} (m^{2} + 1)

. Expanding both sides, we get:

m^{2} r^{2} - 2 m r h + h^{2} = m^{2} r^{2} + r^{2}

The

m^{2} r^{2}

terms on both sides cancel out perfectly. We are left with

- 2 m r h + h^{2} = r^{2}

Rearranging to isolate

m

, we find

2 m r h = h^{2} - r^{2}

, which gives the slope:

m = \frac{h ^{2} - r ^{2}}{2 r h}

Substituting this back into

y = m x

, we get

y = (\frac{h ^{2} - r ^{2}}{2 r h}) x

. Rearranging this, we arrive at the final equation of the second tangent:

(h^{2} - r^{2}) x - 2 r h y = 0

We have conquered the problem by combining geometric intuition with precise algebraic steps.

Common Traps & FAQs

Wait, why is

x = 0

a tangent? I thought I had to solve a quadratic equation to find the tangents.

That is a fantastic question! While the quadratic method works, geometry offers a shortcut. If the distance from the center to a line equals the radius, it *is* a tangent. Since the center is at

(r, h)

and the radius is

r

, the distance to the

y

-axis (

x = 0

) is exactly

r

. It's a tangent by definition!

Why did the

m^{2}

term cancel out when I squared the equation? Did I do something wrong?

You did everything perfectly! It's not a mistake; it's a beautiful simplification. When you square

(m r - h)^{2} = r^{2} (m^{2} + 1)

, the

m^{2} r^{2}

terms appear on both sides. They cancel out, leaving you with a linear equation in

m

. This happens because one of the tangents is vertical (or has a specific slope), and the algebra reflects that.

Can I use the discriminant method (

D = 0

) instead of the distance formula?

Absolutely! You can substitute

y = m x

into the circle equation to get a quadratic in

x

, then set the discriminant to zero. It is mathematically equivalent. However, the distance formula is often faster and less prone to messy algebraic errors. Choose the tool that feels most intuitive to you!

Animated Solution for Mathematics - Circles: The equations of the tangents drawn from the origin to the circle $x^{2} + y^{2} - 2 r x - 2 h y + h^{2} = 0$ , are

Select Answer:

Visualized Solution

Analyze the Circle Equation

Identify Center and Radius

Geometric Observation: The Y-Axis

Equation of the Second Tangent

Condition for Tangency

Squaring and Expanding

Isolating the Slope $m$

Final Equations of Tangents

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Elegant Shortcut

The Algebraic Hunt

The Final Calculation

Similar Questions

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Let $y = x + 2, 4 y = 3 x + 6$ and $3 y = 4 x + 1$ be three tangent lines to the circle $(x - h)^{2} + (y - k)^{2} = r^{2}$ . Then $h + k$ is equal to :

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Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equations of the tangents drawn from the origin to the circle x2+y2−2rx−2hy+h2=0, are

Select Answer:

Visualized Solution

Analyze the Circle Equation

Identify Center and Radius

Geometric Observation: The Y-Axis

Equation of the Second Tangent

Condition for Tangency

Squaring and Expanding

Isolating the Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Final Equations of Tangents

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Elegant Shortcut

The Algebraic Hunt

The Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangents drawn at the point O(0,0) and P(1+5​,2) on the circle x2+y2−2x−4y=0 intersect at the point Q, then the area of the triangle OPQ is equal to

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tangents are drawn to the hyperbola 9x2​−4y2​=1, parallel to the straight line 2x−y=1. The points of contact of the tangents on the hyperbola are

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Animated Solution for Mathematics - Circles: The equations of the tangents drawn from the origin to the circle $x^{2} + y^{2} - 2 r x - 2 h y + h^{2} = 0$ , are

Isolating the Slope $m$

If a tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is

If the tangents drawn at the point $O (0, 0)$ and $P (1 + 5, 2)$ on the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ intersect at the point $Q$ , then the area of the triangle $O P Q$ is equal to

Let $y = x + 2, 4 y = 3 x + 6$ and $3 y = 4 x + 1$ be three tangent lines to the circle $(x - h)^{2} + (y - k)^{2} = r^{2}$ . Then $h + k$ is equal to :

The line $2 x - y + 1 = 0$ is a tangent to the circle at the point $(2, 5)$ and the centre of the circle lies on $x - 2 y = 4$ . Then, the radius of the circle is:

Tangents are drawn to the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1$ , parallel to the straight line $2 x - y = 1$ . The points of contact of the tangents on the hyperbola are

A hyperbola passes through the point $P (2, 3)$ and has foci at $(\pm 2, 0)$ . Then the tangent to this hyperbola at P also passes through the point: