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NVAMediumJEE 2023Circle Equation & Properties

JEE Mathematics 2023 Question with Solution

A circle passing through the point P(α,β)P(\alpha,\beta) in the first quadrant touches the two coordinate axes at the points A and B. The point PP is above the line AB. The point Q on the line segment AB is the foot of the perpendicular from PP on AB. If PQ is equal to 1111 units, then the value of αβ\alpha\beta is:

Answer

Correct answer:121

Step-by-step solution

Standard Method

Given: A circle touches both coordinate axes at points A and B, and passes through P(α,β)P(\alpha,\beta). Also, PQ=11PQ = 11, where Q is the foot of the perpendicular from PP to AB.

Find: The value of αβ\alpha\beta.

If the circle touches both coordinate axes, its centre is C(r,r)C(r,r) and its equation is

(x−r)2+(y−r)2=r2(x-r)^2 + (y-r)^2 = r^2
Circle in the first quadrant touching both coordinate axes at A and B, with point P(alpha, beta), segment AB, perpendicular from P to AB, and length 11 marked.

Since P(α,β)P(\alpha,\beta) lies on the circle,

(α−r)2+(β−r)2=r2(\alpha-r)^2 + (\beta-r)^2 = r^2

Expanding,

α2+β2−2r(α+β)+r2=0\alpha^2 + \beta^2 - 2r(\alpha+\beta) + r^2 = 0

The line joining the points of contact is

x+y=rx + y = r

So the perpendicular distance from P(α,β)P(\alpha,\beta) to AB is

PQ=α+β−r2=11PQ = \frac{\alpha + \beta - r}{\sqrt{2}} = 11

Hence,

α+β=112+r\alpha + \beta = 11\sqrt{2} + r

Now square this relation:

(α+β)2=(112+r)2(\alpha + \beta)^2 = (11\sqrt{2} + r)^2 α2+β2+2αβ=242+r2+222r\alpha^2 + \beta^2 + 2\alpha\beta = 242 + r^2 + 22\sqrt{2}r

From the circle equation,

α2+β2−2r(α+β)+r2=0\alpha^2 + \beta^2 - 2r(\alpha+\beta) + r^2 = 0

Using α+β=112+r\alpha + \beta = 11\sqrt{2} + r,

α2+β2−2r(112+r)+r2=0\alpha^2 + \beta^2 - 2r(11\sqrt{2}+r) + r^2 = 0 α2+β2=222r+r2\alpha^2 + \beta^2 = 22\sqrt{2}r + r^2

Substitute this into

α2+β2+2αβ=242+r2+222r\alpha^2 + \beta^2 + 2\alpha\beta = 242 + r^2 + 22\sqrt{2}r

We get

222r+r2+2αβ=242+r2+222r22\sqrt{2}r + r^2 + 2\alpha\beta = 242 + r^2 + 22\sqrt{2}r 2αβ=2422\alpha\beta = 242 αβ=121\alpha\beta = 121

Therefore, the value of αβ\alpha\beta is 121121.

Coordinate Geometry Expansion

Given: The circle touches both axes and passes through P(α,β)P(\alpha,\beta). The perpendicular distance from PP to AB is 1111.

Find: αβ\alpha\beta.

Let the radius be aa. Then the centre is (a,a)(a,a), so the equation of the circle is

(x−a)2+(y−a)2=a2(x-a)^2 + (y-a)^2 = a^2

Since P(α,β)P(\alpha,\beta) lies on it,

(α−a)2+(β−a)2=a2(\alpha-a)^2 + (\beta-a)^2 = a^2

which gives

α2+β2−2aα−2aβ+a2=0\alpha^2 + \beta^2 - 2a\alpha - 2a\beta + a^2 = 0

The chord joining the touching points with the axes is the line

x+y=ax + y = a

Therefore,

PQ=α+β−a2=11PQ = \frac{\alpha + \beta - a}{\sqrt{2}} = 11

So,

α+β−a=112\alpha + \beta - a = 11\sqrt{2} α+β=a+112\alpha + \beta = a + 11\sqrt{2}

Now,

(α+β)2=a2+242+222a(\alpha + \beta)^2 = a^2 + 242 + 22\sqrt{2}a

that is,

α2+β2+2αβ=a2+242+222a\alpha^2 + \beta^2 + 2\alpha\beta = a^2 + 242 + 22\sqrt{2}a

From the earlier circle relation,

α2+β2=2a(α+β)−a2\alpha^2 + \beta^2 = 2a(\alpha+\beta) - a^2

Using α+β=a+112\alpha + \beta = a + 11\sqrt{2},

α2+β2=2a(a+112)−a2=a2+222a\alpha^2 + \beta^2 = 2a(a + 11\sqrt{2}) - a^2 = a^2 + 22\sqrt{2}a

Hence,

a2+222a+2αβ=a2+242+222aa^2 + 22\sqrt{2}a + 2\alpha\beta = a^2 + 242 + 22\sqrt{2}a

So,

2αβ=2422\alpha\beta = 242 αβ=121\alpha\beta = 121

Therefore, the required value is 121121.

Common mistakes

  • Using the wrong equation of the line AB. Since the circle touches the axes at (r,0)(r,0) and (0,r)(0,r), the line is x+y=rx+y=r, not x−y=rx-y=r. Use the intercept form carefully.

  • Applying the point-to-line distance formula incorrectly. For the line x+y−r=0x+y-r=0, the distance from (α,β)(\alpha,\beta) is ∣α+β−r∣2\frac{|\alpha+\beta-r|}{\sqrt{2}}. Forgetting the denominator 2\sqrt{2} leads to a wrong relation.

  • Expanding (α+β)2(\alpha+\beta)^2 without the middle term. The correct identity is (α+β)2=α2+β2+2αβ(\alpha+\beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta. Missing 2αβ2\alpha\beta prevents extraction of the required product.

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