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MHT-CET Maths · Circle

Concentric Circles and Circles Touching a Line or an Axis

A concentric circle keeps −g and −f and changes only c; a circle touches a line when its radius equals the distance from the centre to that line, and it touches the X-axis when r = |k|.

All Circle notesMHT-CET Maths strategy

Why this matters

6 PYQs at 17% HARD — the chapter's smallest page and its most repetitive. Concentric with a given circle and double its area (twice), concentric and passing through the centre of another circle, concentric and touching the X-axis, centre given and touching a line, and the point of contact of a circle with a line. Every one is 'keep the centre, find r' or 'r is a distance'.

Concept 1 of 2

Concentric Circles: Same Centre, New Radius From Area or From a Point

Intuition

Concentric circles share gg and ff; only the constant changes. Double the area means R2=2r2R^2 = 2r^2; passing through a given point means RR is the distance from the centre to that point.

Definition

  • x2+y2−6x−4y−12=0x^2 + y^2 - 6x - 4y - 12 = 0: centre (3,2)(3, 2), r=5r = 5. Double area: R2=50R^2 = 50: (x−3)2+(y−2)2=50⇒x2+y2−6x−4y=37(x - 3)^2 + (y - 2)^2 = 50 \Rightarrow x^2 + y^2 - 6x - 4y = 37.
  • 2x2+2y2−6x+8y+1=02x^2 + 2y^2 - 6x + 8y + 1 = 0: scale to x2+y2−3x+4y+12=0x^2 + y^2 - 3x + 4y + \tfrac12 = 0, r2=94+4−12=234r^2 = \tfrac94 + 4 - \tfrac12 = \tfrac{23}{4}; double: R2=232R^2 = \tfrac{23}{2}; new constant 254−232=−214\tfrac{25}{4} - \tfrac{23}{2} = -\tfrac{21}{4}: 4x2+4y2−12x+16y−21=04x^2 + 4y^2 - 12x + 16y - 21 = 0.
  • Concentric with 2x2+2y2−8x−12y−9=02x^2 + 2y^2 - 8x - 12y - 9 = 0 (centre (2,3)(2, 3)) and through the centre (−4,−5)(-4, -5) of another circle: R=36+64=10R = \sqrt{36 + 64} = 10: x2+y2−4x−6y−87=0x^2 + y^2 - 4x - 6y - 87 = 0.
  • Scale the leading coefficient to 11 BEFORE reading g,f,cg, f, c; a factor of 22 left in place halves the centre wrongly.

Concentric family

x2+y2+2gx+2fy+λ=0double area: R2=2r2x^2 + y^2 + 2gx + 2fy + \lambda = 0 \qquad \text{double area: } R^2 = 2r^2

Worked example

Find the circle concentric with x2+y2+4x−2y−4=0x^2 + y^2 + 4x - 2y - 4 = 0 and passing through (1,5)(1, 5).
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1CircleMODERATE
The equation of the concentric circle, with circle C1C_1 having equation x2+y2−6x−4y−12=0x^2+y^2-6x-4y-12=0 and having double the area of C1C_1, is

[Q127 · 3rd May 2nd Shift · 2023]

Doubling the radius for double the area

Double AREA means R=2 rR = \sqrt2\, r, R2=2r2R^2 = 2r^2. Doubling rr quadruples the area and gives x2+y2−6x−4y=87x^2 + y^2 - 6x - 4y = 87, which is not on the list — but 5050 (the value of R2R^2, not the constant) is.
Drill 2 more on concentric circles: same centre, new radius from area or from a point

Concept 2 of 2

Touching a Line or an Axis: Radius = Distance From the Centre; Contact Point = Foot of the Perpendicular

Intuition

A line is tangent when its distance from the centre equals the radius. So a circle with a known centre that touches a given line has r=r = that distance, and the point of contact is the foot of the perpendicular from the centre. Touching the XX-axis means r=∣k∣r = |k|, the YY-axis r=∣h∣r = |h|.

Definition

  • Centre (3,4)(3, 4), touching 5x+12y−11=05x + 12y - 11 = 0: r=∣15+48−11∣13=4r = \dfrac{|15 + 48 - 11|}{13} = 4: x2+y2−6x−8y+9=0x^2 + y^2 - 6x - 8y + 9 = 0.
  • Concentric with centre (3,2)(3, 2) and touching the XX-axis: r=2r = 2: x2+y2−6x−4y+9=0x^2 + y^2 - 6x - 4y + 9 = 0.
  • Centre (−1,1)(-1, 1) touching x+2y+4=0x + 2y + 4 = 0: the contact point (h,k)(h, k) is on the line and CPCP is perpendicular to it (slope 22): h+2k=−4h + 2k = -4, 2h−k=−32h - k = -3: (−2,−1)(-2, -1).
  • Touching both axes means ∣h∣=∣k∣=r|h| = |k| = r; in the first quadrant the centre is (r,r)(r, r).

Tangency of a line

∣ah+bk+c∣a2+b2=rtouches the X-axis  ⟺  r=∣k∣\frac{|ah + bk + c|}{\sqrt{a^2 + b^2}} = r \qquad \text{touches the } X\text{-axis} \iff r = |k|

Worked example

Find the circle with centre (2,−3)(2, -3) that touches the line 3x−4y+7=03x - 4y + 7 = 0.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2CircleMODERATE
The equation of the circle which has its centre at the point (3,4)(3,4) and touches the line 5x+12y−11=05x+12y-11=0 is

[Q136 · 4th May Shift 1 · 2023]

Using the centre's x-coordinate for tangency to the X-axis

Touching the XX-axis fixes r=∣k∣r = |k|, the yy-coordinate. With centre (3,2)(3, 2), r=2r = 2, constant 99; using r=3r = 3 gives constant 44, option (D).
Drill 2 more on touching a line or an axis: radius = distance from the centre; contact point = foot of the perpendicular

Summary — formulas & gotchas at a glance

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