Test: Circle - 4 - JEE MCQ

Let r be the radius of required circle. Now, if two circles touches each other then distance between their centres =|r ± 2| = 5 (given)
∴    r = 3, 7
Note: Equation of circles are x² + y² = 9 or x² + y² = 49

(x−2)²+(y+1)² = r²

(3,6). lies on it

⇒ 1+49=r²

⇒ r²=50

⇒ x²+4−4x+y²+1+2y=50.

⇒ x²+y²−4x+2y−45=0

For both lines to be parallel tangent the distance between both lines
should be equal to the diameter of the circle
⇒ 4 = |c₁−c₂|/(1+3)^1/2
⇒∣c₁−c₂∣ = 8

⇒ 1/8 * ∣c₁−c₂∣ = 1

Given:

• A circle passes through point P(α, β) in the first quadrant.

• It touches the x-axis at A and y-axis at B.

• So, A = (r, 0) and B = (0, r), where r is the radius.

• Line AB has equation: y = -x + r.

• P lies above AB, and a perpendicular from P to AB meets it at Q.

• Given: Distance PQ = 11 units.

• Need to find the value of αβ.

Step-by-step Summary:

1. Line AB:

   • Slope of AB = -1.

   • Equation: y = -x + r.

2. Equation of Circle:

   • Center is at (r, r) (since it touches both axes).

   • Equation: (x - r)² + (y - r)² = r²

   • After expanding: x² + y² - 2r(x + y) + r² = 0

   • Substituting point P(α, β) into the equation gives: α² + β² - 2r(α + β) + r² = 0 —— (Equation A)

3. Foot of Perpendicular Q:

   • Perpendicular from P to AB has slope = 1.

   • Equation: y = x + (β - α)

   • Solving with AB gives coordinates of Q as: x = (r - β + α)/2, y = (r + β - α)/2

4. Distance PQ = 11:

   • Using perpendicular distance from P(α, β) to line AB:

   • Distance = |α + β - r| / √2

   • Given PQ = 11, so: |α + β - r| = 11√2

   • Since P is above AB, α + β - r > 0, so: r = α + β - 11√2 —— (Equation B)

5. Substitute r in Equation A:

   • Substitute r = α + β - 11√2 into Equation A.

   • After simplifying, all terms cancel out except -2αβ + 242 = 0

   • Solving gives: αβ = 121

Final Answer:

αβ = 121

Step 1: Analyze Circle C₁

The equation of circle C₁ is:

x² + y² - 4x - 2y = α - 5

To convert it into standard form, complete the square:

• Group x terms: x² - 4x → (x - 2)² - 4

• Group y terms: y² - 2y → (y - 1)² - 1

So, the equation becomes:

(x - 2)² + (y - 1)² = α

This represents a circle with:

• Center at (2, 1)

• Radius = √α

Step 2: Mirror Image of C₁ in the Line y = 2x + 1

We are told that the mirror image of C₁ in the line y = 2x + 1 is another circle C₂, with the equation:

5x² + 5y² - 10fx - 10gy + 36 = 0

Divide the entire equation by 5 to simplify:

x² + y² - 2fx - 2gy + 36/5 = 0

Complete the square:

• x² - 2fx = (x - f)² - f²

• y² - 2gy = (y - g)² - g²

So it becomes:

(x - f)² + (y - g)² = f² + g² - 36/5

This is a circle with:

• Center at (f, g)

• Radius = √(f² + g² - 36/5)

Step 3: Reflect the Center of C₁ Across the Line y = 2x + 1

Rewrite the line as: 2x - y + 1 = 0
This gives: a = 2, b = -1, c = 1

The original center of C₁ is (2, 1)

The formula for the reflection of point (x₀, y₀) across ax + by + c = 0 is:

• x' = x₀ - [2a(ax₀ + by₀ + c)] / (a² + b²)

• y' = y₀ - [2b(ax₀ + by₀ + c)] / (a² + b²)

Calculate step-by-step:

• a² + b² = 4 + 1 = 5

• ax₀ + by₀ + c = 2*2 + (-1)*1 + 1 = 4 - 1 + 1 = 4

Now:

• x' = 2 - (224)/5 = 2 - 16/5 = -6/5

• y' = 1 - (2*(-1)*4)/5 = 1 + 8/5 = 13/5

So, the center of the mirror image C₂ is:

(f, g) = (-6/5, 13/5)

Step 4: Radius of C₂

Using the formula:

Radius r = √(f² + g² - 36/5)

Now substitute:

• f² = (-6/5)² = 36/25

• g² = (13/5)² = 169/25

• f² + g² = 205/25 = 41/5

Then:

• r² = 41/5 - 36/5 = 5/5 = 1

• r = √1 = 1

Step 5: Use Radius to Find α

We earlier found that the radius of circle C₁ is √α
Now we found that the radius of its mirror image is r = 1

Since a reflection doesn’t change radius:

√α = 1 → α = 1

Step 6: Final Answer

We need to find α + r:

α = 1, r = 1

So:

α + r = 1 + 1 = 2
Thus out answer is 2