Equation of two diameters of a circle are 2x-3y=5$2x-3y=5$ and 3x-4y=7$3x-4y=7$. The line joining the points (-(22)/(7),-4)$(-\frac{22}{7},-4)$ and (-(1)/(7),3)$(-\frac{1}{7},3)$intersects the circle at only one pointP(α,β)$P(\alpha,\beta)$. Then 17β-α$17\beta-\alpha$ is equal to
Numerical Answer Type:
Enter a numerical valueAnswer: 2 to 2+4 marks
Solution & Explanation
Related Formula
The intersection of any two non-parallel diameters yields the center of the circle.
A line that intersects a circle at exactly one point is a tangent line.
The tangent at the point of contact P$P$ is always perpendicular to the radius CP$CP$, thus mtangent × mradius = -1$m_{\text{tangent}} \times m_{\text{radius}} = -1$.
Core Logic
Find the center C$C$ by solving the two diameter equations:
Multiply (1) by 3 and (2) by 2:
6x - 9y = 15$6x - 9y = 15$6x - 8y = 14$6x - 8y = 14$
Subtracting the equations gives -y = 1 ⇒ y = -1$-y = 1 \Rightarrow y = -1$.
Substitute y = -1$y = -1$ into (1): 2x + 3 = 5 ⇒ 2x = 2 ⇒ x = 1$2x + 3 = 5 \Rightarrow 2x = 2 \Rightarrow x = 1$.
Center C$C$ is (1, -1)$(1, -1)$.
Find the equation of the line joining A(-(22)/(7), -4)$A(-\frac{22}{7}, -4)$ and B(-(1)/(7), 3)$B(-\frac{1}{7}, 3)$.
Slope of AB$AB$: mAB = (3 - (-4))/(-1/7 - (-22/7)) = (7)/(21/7) = (7)/(3)$m_{AB} = \frac{3 - (-4)}{-1/7 - (-22/7)} = \frac{7}{21/7} = \frac{7}{3}$.
Equation of line AB$AB$:
Since line AB$AB$intersects the circle at only one pointP(α, β)$P(\alpha, \beta)$, line AB$AB$ is a tangent to the circle, and P$P$ is the point of tangency.
The radius line CP$CP$ is perpendicular to tangent AB$AB$.
Slope of CP$CP$ (mCP$m_{CP}$) must be -(3)/(7)$-\frac{3}{7}$.
Equation of the line passing through center C(1, -1)$C(1, -1)$ with slope -(3)/(7)$-\frac{3}{7}$:
When a line "intersects a circle at exactly one point", it's a coded cue to stop thinking about quadratics and discriminants, and immediately build a perpendicular geometric radius from the center to find the exact tangency coordinate.
Chapter Mix
Class 11 Mathematics: Circles
Class 11 Mathematics: Straight Lines
Keywords:#intersects the circle at only one point#JEE Main 2024 Morning Q21#Circles JEE Main 2024#Tangents and Normal JEE Main 2024
More Circles Previous-Year Questions — Page 4
Q8jee_main_2026_23_january_eveningFamily of Curves
If the points of intersection of the ellipsesx² + 2y² - 6x - 12y + 23 = 0$x^{2} + 2y^{2} - 6x - 12y + 23 = 0$ and
4x² + 2y² - 20x - 12y + 35 = 0$4x^2 + 2y^{2} - 20x - 12y + 35 = 0$
lie on a circle of radius r$r$ and centre (a, b)$(a, b)$, then the value of ab + 18r²$ab + 18r^{2}$ is
A.53$53$
B.51$51$
C.52$52$
D.55$55$
Solution
Related Formula
The equation of a family of curves passing through the intersection of two conics S₁ = 0$S_1 = 0$ and S₂ = 0$S_2 = 0$ is S₁ + λ S₂ = 0$S_1 + \lambda S_2 = 0$.
For this resulting curve to be a circle, the coefficient of x²$x^2$ must equal the coefficient of y²$y^2$, and the coefficient of the xy$xy$ term must be zero.
When intersection points of two 2nd$2^{\text{nd}}$ degree curves form a circle, apply S₁ + λ S₂ = 0$S_1 + \lambda S_2 = 0$ immediately, forcing the necessary symmetric coefficients to extract λ$\lambda$.
Chapter Mix
Class 11 Maths: Conic Sections
Class 11 Maths: Circles
Q9jee_main_2026_23_january_eveningProperties of Hyperbola
Let PQ$PQ$ be a chord of the hyperbola x²4- y²b²=1$\frac{x^{2}}{4}-\frac{y^{2}}{b^{2}}=1$, perpendicular to the x-axis such that OPQ$OPQ$ is an equilateral triangle, O$O$ being the centre of the hyperbola. If the eccentricity of the hyperbola is √(3)$\sqrt{3}$, then the area of the triangle OPQ$OPQ$ is:
A.2√(3)$2\sqrt{3}$
B.8√(3)5$\frac{8\sqrt{3}}{5}$
C.(11)/(5)$\frac{11}{5}$
D.(9)/(5)$\frac{9}{5}$
Solution
Related Formula
Eccentricity of a hyperbola (x²)/(a²) - (y²)/(b²) = 1$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is e = √(1 + (b²)/(a²))$e = \sqrt{1 + \frac{b^2}{a^2}}$
Parametric coordinates on a hyperbola are (a θ, b θ)$(a\sec\theta, b\tan\theta)$.
Core Logic
Properties of Hyperbola diagram for Q9 - JEE Main 2026 Evening
Given e = √(3)$e = \sqrt{3}$ and a² = 4$a^2 = 4$ (a=2$a=2$):
The equation is (x²)/(4) - (y²)/(8) = 1$\frac{x^2}{4} - \frac{y^2}{8} = 1$.
Let the point P$P$ on the hyperbola be (2 θ, 2√(2) θ)$(2\sec\theta, 2\sqrt{2}\tan\theta)$.
Since chord PQ$PQ$ is perpendicular to the x-axis, the triangle OPQ$OPQ$ has the x-axis as its altitude OM$OM$. In equilateral OPQ$\triangle OPQ$, half the vertex angle is 30^°$30^\circ$:
A perpendicular chord symmetric about the principal axis automatically splits into two right triangles. Utilizing the parametric coordinates directly defines the ratio of sides matching 30^°$\tan 30^\circ$.
Chapter Mix
Class 11 Maths: Hyperbola
Q19jee_main_2026_23_january_eveningProperties of Parabola
An equilateral triangle OAB$OAB$ is inscribed in the parabola y²=4x$y^{2}=4x$ with the vertex O$O$ at the vertex of the parabola. Then the minimum distance of the circle having AB$AB$ as a diameter from the origin is
A.4(3 - √(3))$4(3 - \sqrt{3})$
B.2(8 - 3√(3))$2(8 - 3\sqrt{3})$
C.4(6 + √(3))$4(6 + \sqrt{3})$
D.2(3 + √(3))$2(3 + \sqrt{3})$
Solution
Related Formula
Parametric form of y² = 4ax$y^2 = 4ax$ is (at², 2at)$(at^2, 2at)$.
Minimum distance from origin to a circle with centre C$C$ and radius r$r$ is |OC - r|$|OC - r|$.
Core Logic
Properties of Parabola diagram for Q19 - JEE Main 2026 Evening
For parabola y² = 4x$y^2 = 4x$, a = 1$a = 1$.
Let the vertices of the equilateral triangle be O(0,0), A(t², 2t), B(t², -2t)$O(0,0), A(t^2, 2t), B(t^2, -2t)$ due to symmetry across the x-axis.
The slope of OA$OA$ makes an angle of 30^°$30^\circ$ with the x-axis.
Vertices are A((2√(3))², 2(2√(3))) = A(12, 4√(3))$A((2\sqrt{3})^2, 2(2\sqrt{3})) = A(12, 4\sqrt{3})$ and B(12, -4√(3))$B(12, -4\sqrt{3})$.
The circle has diameter AB$AB$. Therefore, its centre is the midpoint of AB$AB$, which is C(12, 0)$C(12, 0)$.
The radius R$R$ is half the length of AB$AB$, so R = 4√(3)$R = 4\sqrt{3}$.
The distance from the origin O(0,0)$O(0,0)$ to the centre C(12, 0)$C(12, 0)$ is d = 12$d = 12$.
The minimum distance from the origin to the circle is |d - R|$|d - R|$:
Any polygon inscribed in a standard parabola symmetric across the axis can be defined completely by equating the slope of a vertex from the origin to the respective tangent of the split angle.
Chapter Mix
Class 11 Maths: Parabola
Class 11 Maths: Circles
Q2jee_main_2026_24_january_morningCircles and Centroid Locus
Let a circle of radius 4$4$ pass through the origin O$O$, the points A(-√(3)a, 0)$A(-\sqrt{3}a, 0)$ and B(0, -√(2)b)$B(0, -\sqrt{2}b)$, where a$a$ and b$b$ are real parameters and ab ≠ 0$ab \neq 0$. Then the locus of the centroid of Δ OAB$\Delta OAB$ is a circle of radius Graphical representation of the triangle OAB mapped on the Cartesian plane.
Graphical representation of the triangle OAB mapped on the Cartesian plane.
Since ∠ AOB = 90^°$\angle AOB = 90^\circ$, AB$AB$ is the diameter of the circle passing through O, A, B$O, A, B$.
Therefore, the diameter AB = 2r = 8$AB = 2r = 8$.
This is a circle with r² = (64)/(9)$r^2 = \frac{64}{9}$, so r = (8)/(3)$r = \frac{8}{3}$.
Pattern Recognition
Whenever a circle passes through the origin and points on the axes, the segment joining the axes points is always the diameter. Centroid coordinates easily scale the locus.
Chapter Mix
Class 11 Maths: Conic Sections
Class 11 Maths: Straight Lines
Let each of the two ellipses E₁: x²a²+ y²b²=1,(a>b)$E_{1}:\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,(a>b)$ and E₂: x²A²+ y²B²=1,(A$E_{2}:\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1,(A have eccentricity (4)/(5)$\frac{4}{5}$. Let the lengths of the latus recta of E₁$E_{1}$ and E₂$E_{2}$ be ₁$\ell_{1}$ and ₂$\ell_{2}$, respectively, such that 2 ₁²=9 ₂$2\ell_{1}^{2}=9\ell_{2}$. If the distance between the foci of E₁$E_{1}$ is 8$8$, then the distance between the foci of E₂$E_{2}$ is Graphical mapping defining major and minor axes orientation.
A.(96)/(5)$\frac{96}{5}$
B.(32)/(5)$\frac{32}{5}$
C.(16)/(5)$\frac{16}{5}$
D.(8)/(5)$\frac{8}{5}$
Solution
Related Formula
Eccentricity e = 1 - minor²major²$$\text{Eccentricity } e = \sqrt{1 - \frac{\text{minor}^2}{\text{major}^2}}$$Distance between foci = 2a e (or 2Be if vertical)$$\text{Distance between foci } = 2a e \ \text{(or } 2Be \text{ if vertical)}$$Latus rectum length = 2minor²major$$\text{Latus rectum length } = \frac{2\text{minor}^2}{\text{major}}$$
Core Logic
For E₁$E_1$: a>b ⇒ e = (4)/(5)$a>b \Rightarrow e = \frac{4}{5}$. Distance between foci 2ae = 8 ⇒ a(4/5) = 4 ⇒ a = 5$2ae = 8 \Rightarrow a(4/5) = 4 \Rightarrow a = 5$.
Notice the axis orientation shift: E₁$E_1$ is horizontal, E₂$E_2$ is vertical. Apply latus rectum formula (2A²)/(B)$\frac{2A^2}{B}$ appropriately to avoid formula blind-spots.
Chapter Mix
Class 11 Maths: Conic Sections
More Circles Questions — jee_main_2024_29_jan_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
