Start free

Mathematics · Circles

JEE Main 2024 — 27 January, Shift 2 — Question 28

Consider a circle (x−α)2+(y−β)2=50(x-\alpha)^{2}+(y-\beta)^{2}=50, where α,β>0\alpha, \beta>0. If the circle touches the line y+x=0y+x=0 at the point PP, whose distance from the origin is 424 \sqrt{2} , then (α+β)2(\alpha+\beta)^{2} is equal to.

Answer: 100

Numerical answer — enter this value.

Step-by-step solution

figure

S:(x−α)2+(y−β)2=50S:(x-\alpha)^{2}+(y-\beta)^{2}=50

CP=r\mathrm{CP}=\mathrm{r}

∣α+β2∣=52\left|\frac{\alpha+\beta}{\sqrt{2}}\right|=5 \sqrt{2} ⇒(α+β)2=100\Rightarrow(\alpha+\beta)^{2}=100

Answer key and solution verified before publishing.

Practise Circles

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2024
Paper
27 January, Shift 2
Subject
Mathematics
Chapter
Circles
Topic
Considering a Line or a Point wrt a Circle
Consider a circle (x-α) 2 +(y-β) 2 =50 , where α, β 0 . If the circle… | JEE Main 2024 PYQ with Solution · DhiX AI