Mathematics · Circles

JEE Advanced 2020 — Paper 2 — Question 36

Let OO be the centre of the circle x2+y2=r2x^{2}+y^{2}=r^{2}, where r>52r>\frac{\sqrt{5}}{2}. Suppose PQ is a chord of this circle and the equation of the line passing through PP and QQ is 2x+4y=52 x+4 y=5. If the centre of the circumcircle of the triangle OPQ lies on the line x+2y=4x+2 y=4, then the value of rr is

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

Let R(h,k)R(h, k) be point of intersection of tangents at PP and QQ on x2−y2=r2x^{2}-y^{2}=r^{2}

⇒\Rightarrow \quad equation of chord of contact PQ is xh+yk=r2\mathrm{xh}+\mathrm{yk}=\mathrm{r}^{2}

Which is also 2x+4y=52 x+4 y=5

⇒(h,k)≡(2r25,4r25)\Rightarrow \quad(\mathrm{h}, \mathrm{k}) \equiv\left(\frac{2 \mathrm{r}^{2}}{5}, \frac{4 \mathrm{r}^{2}}{5}\right)

Mid-point of OR is circum-centre of △OPQ\triangle \mathrm{OPQ}

⇒(r25,2r25)\Rightarrow \quad\left(\frac{r^{2}}{5}, \frac{2 r^{2}}{5}\right) lies on x+2y=4⇒r=2x+2 y=4 \Rightarrow r=2

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2020
Paper
Paper 2
Subject
Mathematics
Chapter
Circles
Topic
Chords connected with a Circle
Let O be the centre of the circle x 2 +y 2 =r 2 , where r frac √(5) 2… | JEE Advanced 2020 PYQ with Solution · DhiX AI