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Maths Question 11 – JEE-MAIN 2025

If the four distinct points (4,6), (1,5), (0,0) and (k,3k) lie on a circle of radius r, then 10k+r2 is equal to

Start by using the general equation of a circle, x2+y2+2gx+2fy+c=0, and the fact that one of the given points is the origin to simplify it.

Step 1: Determine the equation of the circle✦ Active

The general equation of a circle is x2+y2+2gx+2fy+c=0. Since the point (0,0) lies on the circle, substituting it gives c=0. Thus, the equation simplifies to x2+y2+2gx+2fy=0.

Substitute the point (4,6): 42+62+2g(4)+2f(6)=016+36+8g+12f=08g+12f=522g+3f=13(1).

Substitute the point (1,5): (1)2+52+2g(1)+2f(5)=01+252g+10f=02g+10f=26g+5f=13(2).

Solve the system of equations (1) and (2). From (2), g=5f+13. Substitute this into (1): 2(5f+13)+3f=1310f+26+3f=1313f=39f=3. Substitute f=3 back into g=5f+13: g=5(3)+13=15+13=2.

The equation of the circle is x2+y24x6y=0.

Step 2: Calculate r2 and find k○ Expand

The radius squared is r2=g2+f2c. Since c=0, r2=(2)2+(3)2=4+9=13.

The fourth point (k,3k) lies on the circle. Substitute it into the circle's equation: k2+(3k)24k6(3k)=0.

k2+9k24k18k=010k222k=0.

Factor out 2k: 2k(5k11)=0. Since the four points are distinct, k0 (otherwise (k,3k) would be (0,0)). Therefore, 5k11=0k=115.

Step 3: Calculate the final expression○ Expand

We need to find the value of 10k+r2.

Substitute the values of k and r2: 10(115)+13=2×11+13=22+13=35.

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